Question

The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years but their mean SAT score is lower than previous years. He took a sample of 20 of this year's entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university's record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. True or False: The null hypothesis would be rejected.

Answer 1

False.

The null hypothesis failed to be rejected.

At a significance level of 5%, there is not enough evidence to support the claim that the entering class has a mean SAT score that is significantly lower than 1520.

Explanation:

This is a hypothesis test for the population mean.

The claim is that the entering class has a mean SAT score that is significantly lower than 1520.

Then, the null and alternative hypothesis are:

`H_0: \mu=1520\\\\H_a:\mu< 1520`

The significance level is 0.05.

The sample has a size n=20.

The sample mean is M=1501.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=53.

The estimated standard error of the mean is computed using the formula:

`s_M=(s)/(√(n))=(53)/(√(20))=11.851`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(1501-1520)/(11.851)=(-19)/(11.851)=-1.6`

The degrees of freedom for this sample size are:

`df=n-1=20-1=19`

This test is a left-tailed test, with 19 degrees of freedom and t=-1.6, so the P-value for this test is calculated as (using a t-table):

`\text{P-value}=P(t<-1.6)=0.063`

As the P-value (0.063) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the entering class has a mean SAT score that is significantly lower than 1520.