Question

A professor of statistics wants to test that the average amount of money a typical college student spends per day during spring break is over $70. Based upon previous research, the population standard deviation is estimated to be $17.32. The professor surveys 35 students and finds that the mean spending is $72.43. Which of the following statements is most accurate? Select one: a. reject the null hypothesis at ? = 0.01 b. reject the null hypothesis at ? = 0.10 c. fail to reject the null hypothesis at ? ? 0.10 d. reject the null hypothesis at ? = 0.05

Answer 1

Answer:

Option c: Fail to reject the null hypothesis at α = 0.10

Step-by-step explanation:

We are given;

Population mean; μ = 70

population standard deviation; σ = 17.32

Sample mean; x¯ = 72.43

Sample size; n = 35

Thus;

Null hypothesis; H0: μ = 70

Alternative hypothesis; Ha: μ > 70

Z-score formula is;

z = (x¯ - μ)/(σ/√n)

z = (72.43 - 70)/(17.32/√35)

z = 2.43/2.9276

z = 0.83

From online p-value from z-score calculator attached, using z = 0.83; significance level = 0.05 and one tailed hypothesis, we have;

P-value ≈ 0.2033

It's more than the significance value, so we will fail to reject the null hypothesis at α = 0.05

Similarly, From online p-value from z-score calculator attached, using z = 0.83; significance level = 0.01 and one tailed hypothesis, we have;

P-value ≈ 0.2033

It's more than the significance value, so we will fail to reject the null hypothesis at α = 0.01

Similarly, From online p-value from z-score calculator attached, using z = 0.83; significance level = 0.10 and one tailed hypothesis, we have;

P-value ≈ 0.2033

It's more than the significance value, so we will fail to reject the null hypothesis at α = 0.10

From the 3 significance values, the correct option is option c. fail to reject the null hypothesis at α = 0.10

Answer 2

Using the z-distribution, it is found that the most accurate statement is:

• c. fail to reject the null hypothesis at $\alpha = 0.1$

At the null hypothesis, it is tested if the average spending is not over $70, that is:

$H_0: \mu \leq 70$

At the alternative hypothesis, it is tested if it is over $70, that is:

$H_1: \mu > 70$.

We have the standard deviation for the population, hence, the z-distribution is used to solve this question.

The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• $\sigma$ is the standard deviation of the sample.

• n is the sample size.

In this problem, the values of the parameters are as follows: $\overline{x} = 72.43, \mu = 70, \sigma = 17.32, n = 35$.

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{72.43 - 70}{\frac{17.32}{\sqrt{35}}}$

$z = 0.83$

Using a significance level of $\alpha = 0.1$, the critical value for the right-tailed test, as we are testing if the mean is greater than a value, is $z^{\ast} = 1.28$.

• The lower the significance level, the higher the critical value.

Since for a significance level of 0.1, $z < z^{\ast}$, we fail to reject the null hypothesis at $\alpha = 0.1$, and option c is correct.

You can learn more about the use of the z-distribution to test an hypothesis at https://brainly.com/question/16313918