High School

Assume you are testing the claim that the mean age of the prison population in one city is less than 25 years. The sample data are summarized as follows:

- Sample size, [tex]n = 25[/tex]
- Sample mean, [tex]\overline{x} = 24.4[/tex] years
- Sample standard deviation, [tex]s = 9.2[/tex] years

Use a significance level of [tex]\alpha = 0.05[/tex]. Answer the following questions:

1. Calculate the test statistic. Round your answer to three decimal places. (Show all your work for setting up and calculating the test statistic. Calculator-only work will not be accepted.)

2. Determine the p-value and make a decision to reject or fail to reject the null hypothesis. Round each value to three decimal places. (You may use Excel or a calculator to find the p-value, but write down what you input into Excel/calculator.)

3. Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, a Student's t-distribution, or neither. Choose the correct distribution for each of the following claims:

a. Claim: [tex]\mu = 107[/tex]. Sample data: [tex]n = 17[/tex], [tex]\overline{x} = 101[/tex], [tex]s = 15.1[/tex]. The sample data appear to come from a normally distributed population with unknown [tex]\mu[/tex] and [tex]\sigma[/tex].

b. Claim: [tex]\mu = 981[/tex]. Sample data: [tex]n = 23[/tex], [tex]\overline{x} = 912[/tex], [tex]s = 30[/tex]. The sample data appear to come from a normally distributed population with [tex]\sigma = 30[/tex].

4. Assume that you want to test the claim that the paired sample data (matched pairs) come from a population for which the mean difference is [tex]d = 0[/tex]. Compute the value of the test statistic. Round the test statistic to four decimal places. Use the following data:

- [tex]x[/tex]: 28, 31, 20, 25, 26, 27, 33, 35
- [tex]y[/tex]: 26, 27, 26, 25, 29, 32, 33, 34

5. A newspaper article about the results of a poll states: "In theory, the results of such a poll, in 99 cases out of 100, should differ by no more than 5 percentage points in either direction from what would have been obtained by interviewing all voters in the United States." Find the sample size suggested by this statement.

- Shade and label the missing parts in the graph for a 99% confidence interval. (You may use Excel or a calculator to find the z-score. Round the z-score to two decimal places.)

- Area to the left:
- Area in the middle:
- Area to the right:
- [tex]Z = [/tex]
- The sample size is:

Answer :

1. The calculated test statistic is approximately -0.326.

2. The p-value is about 0.373, so we fail to reject the null hypothesis.

3a) The hypothesis test involves a normal distribution and uses the t-distribution.

3b) The test uses the z-distribution for a normally distributed population with a known standard deviation.

4. The computed test statistic for the paired sample data is approximately -0.4862.

5. For a 99% confidence interval, the Z-score is approximately -2.576. Unfortunately, I can't interpret or draw missing parts of a graph without more specific details.

1. Given data:

Sample size (n) = 25

Sample mean (X) = 24.4 years

Sample standard deviation (s) = 9.2 years

Significance level (α) = 0.05

The null hypothesis (H₀) states that the mean age of the prison population is 25 years or more. The alternative hypothesis (H₁) states that the mean age is less than 25 years.

The test statistic for a one-sample t-test is given by:

t= X-μ₀/s/√n

Where:

X is the sample mean

μ₀ is the hypothesized population mean under the null hypothesis (25 years in this case)

s is the sample standard deviation

n is the sample size

Substituting the values:

t = 24.4-25/9.2/√25

Calculating the numerator and denominator separately:

t = -0.6/1.84

​

Now calculating the test statistic:

t≈−0.326

Rounded to 3 decimal places, the test statistic is approximately -0.326.

2. Since this is a one-tailed test (claiming that the mean age is less than 25), we need to find the p-value corresponding to the test statistic -0.326. You can use a t-distribution table, a calculator, or statistical software to find the p-value.

Using the t-distribution with degrees of freedom df = n-1 = 25-1 = 24, the p-value for t=−0.326 is approximately 0.373.

Since the p-value (0.373) is greater than the significance level (α = 0.05), we fail to reject the null hypothesis. There is not enough evidence to conclude that the mean age of the prison population is less than 25 years.

3. a. For a claim involving a normally distributed population with unknown mean and standard deviation, we use the t-distribution.

b. Similarly, for the second claim involving a normally distributed population with known standard deviation (σ = 30), we use the z-distribution.

4. Given paired sample data:

X:28,31,20,25,26,27,33,35

Y:26,27,26,25,29,32,33,34

The paired difference d = X - Y for each pair:

d:2,4,−6,0,−3,−5,0,1

Mean of the differences ([tex]\bar d[/tex]) = (2 + 4 - 6 + 0 - 3 - 5 + 0 + 1) / 8 = -1.125

Standard deviation of the differences (sd) = 3.940

Sample size (n) = 8

Hypothesized population mean difference (d₀) = 0

The test statistic for a paired t-test is given by:

t= d-d₀/sd/√n

Substituting the values:

t=-1.125-0/3.940/√8

t≈−0.4862

Rounded to four decimal places, the test statistic is approximately -0.4862.

5. For a 99% confidence interval, the critical z-score is the z-score corresponding to the middle 99% of the standard normal distribution. This can be found using a standard normal distribution table or calculator.

The area in the middle (between -Z and Z) for a 99% confidence interval is 0.99.

So, the area to the left (and to the right) is (1 - 0.99) / 2 = 0.005.

The Z-score for a cumulative area of 0.005 is approximately -2.576.

For a 99% confidence interval, the Z-score is -2.576.

To know more about test statistic:

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