Answer to Question #305269 in Statistics and Probability for lisa

Question

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 P-value? Should the null hypothesis be rejected at alpha = .05 level of significance?

Answer 1

$H_0:a=a_0=10337$

$H1:a>a_0$

Test statistic:

$T={\frac {(x-a)*\sqrt{n}} {\sigma}}$ , where x - sample mean, n - sample size, $\sigma$ - population standard deviation. So, in the given case we have

$T={\frac {(10798-10337)*\sqrt{150}} {1560}}\approx 3.62$

Since population standard deviation is known and sample size is big, then it is aapropriate to use z-statisctic. So, p-value for obtained T is $p=P(Z>T)=P(Z>3.62)=0.00015$

p < 0.05, so we should reject the null hypothesis and conclude that average expenditure is greater that 10337