Solution to The average expenditure per student (based on average daily attendance) for a certain school year … - Sikademy
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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. Find the 95% confidence level? Should the null hypothesis be rejected?

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The critical value for \alpha = 0.05 is z_c = z_{1-\alpha/2} = 1.96. The corresponding confidence interval is computed as shown below:

CI=(\bar{X}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{\sigma}{\sqrt{n}})

=(10798-1.96\times\dfrac{1560}{\sqrt{150}}, 10798+1.96\times\dfrac{1560}{\sqrt{150}})

=(10548.35, 11047.65)

Therefore, based on the data provided, the 95% confidence interval for the population mean is 10548.35 < \mu < 11047.65, which indicates that we are 95% confident that the true population mean \mu is contained by the interval (10548.35, 11047.65).

The following null and alternative hypotheses need to be tested:



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 \alpha = 0.05, and the critical value for a two-tailed test is z_c = 1.96.

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

The z-statistic is computed as follows:


Since it is observed that |z| = 3.619 > 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, and since p=0.000295<0.05=\alpha, it is concluded that the null hypothesis is rejected.

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

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Question ID: mtid-4-stid-46-sqid-1624-qpid-94