Question

At a local university, a sample of 49 evening students was selected in order to determine whether the average age of the evening students is significantly different from 21. The average age of the students in the sample was 23.25 years. The population standard deviation is known to be 7 years. Determine whether or not the average age of the evening students is significantly different from 21. Use a .1 level of significance.

Calculate the critical value.

I know the null and alternative hypothesis is:

H(0): doesn't equal 21

H(a): equals 21

The test statistic:

z = 23.25-21 / 7/ square root of 49

= 2.25 / 7/7

= 2.25 / 1

= 2.25 ( if you could check this)

The p-value I made an educated guess of 0.01224

Answer 1

Given that, sample size = 49

sample mean = 23.25 and population standard deviation = 7

The null and alternative hypotheses are,

H0 : μ = 21

Ha : μ ≠ 21

This test is two-tailed test.

critical values at 0.1 level of significance are, Z^* = ± 1.645

=> Critical values are -1.645, 1.645

Test statistic is,

=> Test statistic = 2.25

p-value = 2 * P(Z > 2.25) = 2 * 0.0122 = 0.0244

=> p-value = 0.0244

Answer 2

To determine whether the average age of the evening students is significantly different from 21, we can perform a hypothesis test using the given information.

The null hypothesis (H0): The average age of the evening students is equal to 21. The alternative hypothesis (Ha): The average age of the evening students is not equal to 21.

To calculate the critical value at a significance level of 0.1, we need to find the corresponding z-value. The critical z-value can be found using a z-table or a statistical software. In this case, since it is a two-tailed test, we divide the significance level by 2 to get 0.05 for each tail.

Using a standard normal distribution table or software, the critical z-value for a significance level of 0.05 (in each tail) is approximately 1.645.

Now, let's calculate the test statistic: z = (sample mean - population mean) / (population standard deviation / √sample size) = (23.25 - 21) / (7 / √49) = 2.25 / (7 / 7) = 2.25 / 1 = 2.25

Based on the calculated test statistic (z = 2.25), we can compare it to the critical z-value to make a decision.

Since the calculated test statistic (2.25) is greater than the critical z-value (1.645), we can reject the null hypothesis. This means that there is sufficient evidence to conclude that the average age of the evening students is significantly different from 21 at a significance level of 0.1.

Regarding the p-value, you made an educated guess of 0.01224. However, without the exact distribution, it is not possible to determine the precise p-value. The p-value represents the probability of observing a test statistic as extreme as the one calculated (or more extreme) under the null hypothesis. To obtain the exact p-value, you would need to consult a z-table or use statistical software.