Question #284595

Student Expenditures The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10,798. Do these results indicate that the average expenditure has changed? Choose your own level of significance.


1
Expert's answer
2022-01-04T11:28:43-0500

The following null and alternative hypotheses need to be tested:

H0:μ=10337H_0:\mu=10337

H1:μ10337H_1:\mu\not=10337

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is α=0.05,\alpha = 0.05, and the critical value for a two-tailed test is zc=1.96.z_c = 1.96.

The rejection region for this two-tailed test is R={z:z>1.96}.R = \{z: |z| > 1.96\}.

The z-statistic is computed as follows:


z=xˉμσ/n=10798103371560/150z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{10798−10337}{1560/\sqrt{150}}

3.6193\approx3.6193

Since it is observed that z=3.6193>1.96=zc,|z| = 3.6193 >1.96= z_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=2P(Z>3.6193)0.000295,p=2P(Z>3.6193)\approx 0.000295, and since p=0.000295<0.05=α,p= 0.000295<0.05=\alpha, 0.05 it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu is different than 10337,10337, at the α=0.05\alpha = 0.05 significance level.

Therefore, there is enough evidence to claim that the average expenditure has changed, at the α=0.05\alpha = 0.05 significance level.


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