Pre-Calculus Assignment — Book 8.4

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Statistics Syllabus — Chapter 7

Statistics

Chapter 7

1. Book 7.1 page 331: 11 – 28

2. Book 7.2.1 page 346: 31 – 36

3. Book 7.2.2 page 346: 37 – 42

4. Book 7.3.1 page 357: 23 – 28, 37, 38

5. Book 7.3.2 page 358: 29 – 36

6. Book 7.4 page 364: 9 – 14

7. Handout 7410

8. Book 7.5 page 374: 17 – 26

9. Book 7.R page 382: 21, 22, 33, 34, 35, 36, 45, 46, 55, 56

10. Book — Putting it all together page 386

11. Book Chapter Quiz page 387

12. Handout 7R10

Test

Statistics Notes — The Chi-Square Distribution

The Chi-Square Distribution

The graph is asymmetrical, skewed to the right.

All values are non-negative.

Tests of the variance requires us to assume that the population has normally distributed values.

Statistics Notes — Hypothesis Testing for Variance and Standard Deviation

In real life, it is often important to produce consistent predictable results. For instance, companies mass produce thousands of products such as light bulbs, car tires, golf balls… There is very low tolerance for variation.

If the population is normal, we can test the variance and standard deviation of the process using the chi-square distribution with n – 1 degrees of freedom.

To test a variance or a standard deviation of a population that is normally distributed, we can use the χ²–test.

The χ²–test for a variance or standard deviation is not as robust as the tests for the population mean or the population proportion. “Not as robust” meaning it is affected greater by outliers and can thus be less precise than wanted.

So it is essential that when performing the χ²–test for a variance or standard deviation that the population is normally distributed. The results can be misleading if the population is not normal.

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Example. A dairy processing company claims that the variance of the amount of fat in the whole milk processed by the company is no more than 0.25. You suspect this is wrong and find that a random sample of 41 milk containers has a variance of 0.27. At α = 0.05, is there enough evidence to reject the company’s claim? Assume the population is normally distributed.

Claim → σ² ≤ 0.25 This is the null hypothesis, H₀.
H_a: σ² > 0.25. Right-tail test.

n = 41
d.f. = 40
s² = 0.27
σ² = 0.25
α = 0.05

I am going to show you two ways.

1. Rejection regions and find χ².

Our critical value is 55.758. Since it is a right-tailed test, our rejection region is greater than 55.758.

Find χ².

Since 43.2 is not inside our rejection region, we will fail to reject the null hypothesis. There is not enough evidence to reject the company’s claim at the 5% level of significance.

2. Find χ², then find P-value.

Find χ².

Find P-value.

So our P-value is 0.3362 which is more than α = 0.05. Thus we will fail to reject the null. Recall: If P > α, then fail to reject H₀.

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Example. A police chief claims that the standard deviation in the length of response times is less than 3.7 minutes. A random sample of nine response times has a standard deviation of 3.0 minutes. At α = 0.05, is there enough evidence to support the police chief’s claim? Assume the population is normally distributed.

Claim → σ < 3.7 This is the alternate hypothesis, H_a. Left-tail test.
H₀: σ ≥ 3.7

n = 9
d.f. = 8
s = 3.0
s² = 9
σ = 3.7
σ² = 13.69
α = 0.05

I am going to show you two ways.

1. Rejection regions and find χ².

Why are we using right tail? This is the way the chi-square table for your text is designed. So if its left-tail 5%, use the number for right-tail 95%.

So our critical value is 2.733. Since it is a left-tailed test, our rejection region will be less than 2.733.

Find χ².

Since 5.2593 is not less than 2.733, it is not in our rejection region. Therefore, we will fail to reject the null.

2. Find χ², then find P-value.

Find χ².

Find P-value.

When using χ²cdf, the -10000 can be replaced with 0 in forming an interval going to the extreme left. You can use the -10000, as you can see, in forming this interval. This would keep the formation of those intervals consistent no matter which cdf function we use.

So our P-value is 0.2705 which is more than α = 0.05. Thus we will fail to reject the null.

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Example. A tire manufacturer claims that the variance of the diameters in a certain tire model is 8.6. A random sample of ten tires has a variance of 4.3. At α = 0.01, is there enough evidence to reject the manufacturer’s claim? Assume the population is normally distributed.

Claim → σ² = 8.6 This is the null hypothesis, H₀.
H_a: σ² ≠ 8.6
Two-tailed test.

n = 10
d.f. = 9
s² = 4.3
σ² = 8.6
α = 0.01

I am going to show you two ways.

1. Rejection regions and find χ².

Our critical values are 1.735 and 23.589. Our rejection regions are less than 1.735 and more than 23.589. Any values between 1.735 and 23.589 are not rejected.

Find χ².

Since 4.5 is not less than 1.735 nor greater than 23.589, it does not fall within our rejection regions. Thus we will fail to reject the null.

2. Find χ², then find P-value.

Find χ².

Find P-value.

So our P-value is more than α = 0.01. Thus we will fail to reject the null.

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Example. The mean lifetime of electric light bulbs produced by a company has in the past been 1120 hours with a standard deviation of no more than 125 hours. A sample of 40 bulbs revealed a standard deviation in the lifetimes of 150 hours. Would you conclude that this is unusual?

Claim → σ ≤ 125 This is the null hypothesis, H₀.
H_a: σ > 125
Right-tail test.

n = 40
d.f. = 39
s = 150
s² = 22500
σ = 125
σ² = 15625
α = not given

Find χ².

Find P-value.

How do we answers this question? Remember, to use a P-value to make a conclusion in a hypothesis test, compare the P-value to α.

1. If P ≤ α, then reject H₀.

2. If P > α, then fail to reject H₀.

P = 0.03693

So at a 1% level of significance, α = 0.01, we would fail to reject the null since P > α. This doesn’t mean we accept the null and thus the claim unfortunately.

At a 5% level of significance, α = 0.05, we would reject the null and thus reject the claim since P ≤ α. These 40 light bulbs would be considered unusual. Marvelous.

Statistics Notes — Hypothesis Testing for Population Proportions

z-Test for a proportion p

Given a binomial distribution such that np ≥ 5 and nq ≥ 5, you can use a z-test to test a population proportion p.

Example. A medical researcher claims that less than 20% of adults in the US are allergic to a medication. In a random sample of 100 adults, 15% say they have such an allergy. Test the researcher’s claim at α = 0.01.

claim → p < 0.20 This is the alternate hypothesis, H_a. Left-tail test.
H₀: p ≥ 0.20

n = 100
ÿ = 0.15
p = .20
q = .80
np ≥ 5 (yes, 100 × .20 = 20)
nq ≥ 5 (yes, 100 × .80 = 80)
α = 0.01

I am going to show you three ways.

1. Rejection regions and z-score.

The critical z value is –2.326. Remember, it is negative because it is a left-tail test.

Find z.

Our z-score is –1.25 which is not inside the rejection region, therefore we will fail to reject the null hypothesis. There is not enough evidence to support the claim that less than 20% of adults in the US are allergic to the medication.

2. Find z-score, then P-value.

Find z.

Find P-value.

Since P > α, then we will fail to reject the null hypothesis. This does not mean we accept the null however. We do not have enough evidence to do accept the null hypothesis.

3. Using that fun fun graphing calculator. YAY.

For x, use n × ÿ.

Since P > α, then we will fail to reject the null hypothesis. This does not mean we accept the null however. We do not have enough evidence to do accept the null hypothesis.

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Example. In screening women for breast cancer, doctors use a method that fails to detect cancer in 20% of the women who actually have the disease. Suppose a new method has been developed that researchers hope will detect cancer more accurately. This new method was used to screen a random sample of 140 women known to have breast cancer. Of these, the new method failed to detect cancer in 12 women. Does this sample provide evidence at α = 0.05 that the failure rate of the new method differs from the one currently in use?

claim → p = 0.20 This is the null hypothesis, H₀.
H_a: p ≠ 0.20
Two-tailed test.

n = 140
ÿ = 12/140 = .0857
p = .20
q = .80
np ≥ 5 (yes, 140 × .20 = 28)
nq ≥ 5 (yes, 140 × .80 = 112)
α = 0.05

I am going to this one three ways too.

1. Rejection regions and z-score.

The critical z value is ±1.960. Remember, it is positive or negative because it is a two-tail test.

Find z.

Our z-score is –3.3810 which is inside our rejection regions, thus we will reject the null hypothesis. The rate of detection for the new test differs from the old at a .05 level of significance.

2. Find z-score, then P-value.

Find z.

Find P-value.

Our P-value is 0.0007223. Since P ≤ α, we will reject the null hypothesis.

3. Super fun graphing utility.

Our P-value is 0.0007223. Since P ≤ α, we will reject the null hypothesis.

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Example. The popularity of personal watercraft (also known as jet skis) continues to increase, despite the apparent danger associated with their use. A sample of 54 watercraft accidents reported to the Nebraska Game and Parks Commission in 1997 revealed that 85% of them involved personal watercraft even though only 8% of the motorized boats registered in the state are personal watercraft. Suppose the national average proportion of watercraft accidents in 1997 involving personal watercraft was 78%. Does the watercraft accident rate in Nebraska exceed the rate in the nation? Use the 0.01 level of significance.

claim → p > 0.78
This is the alternate hypothesis, H_a. Right-tailed test.
H₀: p ≤ 0.78

n = 54
ÿ = .85
p = .78
q = .22
np ≥ 5 (yes, 54 × .78 = 42.12)
nq ≥ 5 (yes, 54 × .22 = 11.88)
α = 0.01

I am going to this one in two of the ways – numbers 1 and 3.

1. Rejection regions and z-score.

The critical z value is 2.326.

Find z.

Our z-score is 1.2418 which is not inside our rejection region, thus we will fail to reject the null hypothesis. There is not enough evidence to support the claim that the watercraft accident rate in Nebraska exceed the rate in the nation.

3. Yay, that’s right, its graphing calculator menu fun time.

Since P > α, then we will fail to reject the null hypothesis. There are a few differences in this last screen when compared to prior found values. Recall that x equals n × ÿ. This is 45.9. But if we try and use this for the x, we will get an error screen. So by rounding to 46, this small amount, some of the values will change. C’est la vie.

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Example. In 1990, 5.8% of job applicants who were tested for drugs failed the test. At the 0.01 level of significance test the claim that the failure rate is now lower if a random sample of 1520 current job applicants results in 58 failures.

claim → p < 0.058 This is the alternate hypothesis, H_a. Left-tailed test.
H₀: p ≥ 0.058

n = 1520
ÿ = 58/1520 = .0382
p = .058
q = .942
np ≥ 5 (yes, 1520 × .058 = 88.61)
nq ≥ 5 (yes, 1520 × .942 = 1431.84)
α = 0.01

I am going to this one in only one way.

1. Rejection regions and z-score.

The critical z value is –2.326.

Find z.

Our z-score is –3.3025 which is inside our rejection region, thus we reject the null hypothesis. There is enough evidence at a 0.01 level to support the claim that the failure rate is now lower.

Statistics Notes — Hypothesis Testing for the Mean (Small Samples)

It is often not practical to collect samples of size 30 or more.
However, if the population has a normal, or nearly normal, distribution, we can still test the population mean.
To do so, we will use the t-sampling distribution with n – 1 degrees of freedom.

The t-test is a statistical test like the z-test that is used to determine if the difference between a sample mean and the population mean is large enough to be statistically significant, that is, if it is unlikely to have occurred by chance.

Example. A major car manufacturer wants to test a new engine to determine whether it meets new air-pollution standards. The mean emission for all engines of this type must be less than 20 parts per million of carbon. Ten engines are manufactured for testing purposes, and the emission level for each is determined. The mean and standard deviation for the tests are 17.17 and 2.98 parts per million of carbon. Does the data supply enough evidence at α = 0.01 to allow the manufacturer to conclude that this type of engine meets the pollution standard?

claim → μ < 20
The claim is the alternate hypothesis, H_a.
Left-tail test
H₀: μ ≥ 20

n = 10
d.f. = 9
˜ = 17.17
s = 2.98
α = 0.01

I am going to show you three ways of doing this.

1. Rejection regions and t-score.

The critical value for this left-tail test is t_c = –2.821. This critical value is negative because it is a left-tail critical value.

Find t.

Our t-score is –3.0031 which is inside the rejection region, therefore we will reject the null hypothesis thus allowing the manufacturer to conclude that this type of engine meets the pollution standards. Yay. Go Green.

2. Find t-score, then P-value.

Find t.

Find P-value. Remember when using the tcdf command, we type the interval followed by the degrees of freedom.

Since P ≤ α, we will reject the null hypothesis allowing the manufacturer to conclude that this type of engine meets the pollution standards. Yay. Go Green again.

3. That wonderful graphing utility.

Since P ≤ α, we will reject the null hypothesis allowing the manufacturer to conclude that this type of engine meets the pollution standards. Yay. Go Green again again.

Example. Most water treatment facilities monitor the quality of their drinking water on an hourly basis. One variable monitored it is pH level, which measures the degree of alkalinity or acidity in the water. A pH below 7.0 is acidic, one above 7.0 is alkaline, and a pH of 7.0 is neutral. One water treatment plant has a target pH of 8.5 (most try to maintain a slightly alkaline level). Based on 17 water samples at this plant, the mean of one hour’s test results is 8.42 and the standard deviation is 0.16. Does this sample provide sufficient evidence at α = 0.05 that the mean pH level in the water differs from 8.5?

claim → μ = 8.5
The claim is the null hypothesis, H₀.
H_a: μ≠8.5
Two-tailed test

n = 17
d.f. = 16
˜ = 8.42
s = 0.16
α = 0.05

I am going to show you all three ways again.

1. Rejection regions and t-score.

The critical values are ±2.120. Positive or negative because it is a two-tail test.

Find t-score.

Our t-score is –2.0616 which is not inside the rejection regions, therefore we will fail to reject the null hypothesis. This does not mean we accept the null however. We do not have enough evidence to do accept the null hypothesis.

2. Find t-score, then P-value.

Find t-score.

Find P-value.

Recall in a two-tail test we double this first probabilty. It is the left tail. So we double so we have the left and right tails.

Since P > α, then we will fail to reject the null hypothesis. This does not mean we accept the null however. We do not have enough evidence to do accept the null hypothesis.

3. That wonderful graphing utility with its fun menus.

Since P > α, then we will fail to reject the null hypothesis. This does not mean we accept the null however. We do not have enough evidence to do accept the null hypothesis. Yay H2o.

The concept of pH was first introduced by Danish chemist Søren Peder Lauritz Sørensen at the Carlsberg Laboratory in 1909.
pH is a measure of the acidity or basicity of a solution. It is an expression for the effective concentration of hydrogen ions in a solution. Pure water is said to be neutral. The pH for pure water is close to 7.0. Solutions with a pH less than 7 are said to be acidic and solutions with a pH greater than 7 are said to be basic or alkaline. pH measurements are important in medicine, biology, chemistry, food science, environmental science, oceanography and many other applications, such as this statistics example.

Pre-Calculus — Extra Credit Videos

Statistics Assignment — Book 7.2