Question

Use technology and a​ t-test to test the claim about the population mean µ at the given level of significance α using the given sample statistics. Assume the population is normally distributed. Claim: µ > 79​; α = 0.05 Sample​ statistics: xbar = 79.4​, s = 3.3​, n = 23. What are the null and alternative​ hypotheses? What is the value of the standardized test​ statistic?

Answer 1

The null hypothesis is H0: μ ≤ 79, and the alternative hypothesis is Ha: μ > 79. The standardized test statistic for the one-tailed t-test with a sample mean of 79.4, a sample standard deviation of 3.3, and a sample size of 23 is approximately 0.5806.

Step-by-step explanation:

To test the claim about the population mean μ with the level of significance α provided, the null hypothesis (H0) and alternative hypothesis (Ha) need to be set up for a one-tailed t-test. Given the claim that μ > 79 and α = 0.05, the hypotheses are as follows:

• H0: μ ≤ 79
• Ha: μ > 79

The standardized test statistic can be calculated using the t-distribution, since the standard deviation of the population is unknown and the sample size is less than 30.

With a sample mean (xbar) of 79.4, a sample standard deviation (s) of 3.3, and a sample size (n) of 23, the t-statistic (t) is determined using the formula:

t = (xbar - μ) / (s / √n)

Plugging in the values gives:

t = (79.4 - 79) / (3.3 / √23)

t = 0.4 / (3.3 / 4.7958)

t ≈ 0.4 / 0.6887

t ≈ 0.5806

To find the critical value and make a decision about the hypothesis, we would look at a t-distribution table or use technology such as a calculator or statistical software with the degrees of freedom df = n - 1 which is 22.

We compare our calculated t statistic with the critical t-value at the 0.05 significance level for a right-tailed test.