In 2017, the average credit score for loans that were purchased by a government-sponsored mortgage loan company was 744. A random sample of 35 mortgages recently purchased by the company was selected, and it was found that the average credit score was 753 with a sample standard deviation of 19. Complete parts a through c.

a. Using a = 0.05, is there enough evidence from this sample to conclude that the average credit score for mortgages purchased by the company has increased since 2017?
Determine the null and alternative hypotheses. Hos 744 H: > 744 (Type integers or decimals. Do not round.) Determine the appropriate critical value. Select the correct choice below and fill in the answer box within your choice. (Round to three decimal places as needed.)
A. -ta=
B. tą = 1.691
C. ta/2=___-

b. Determine the precise p-value for this test using Excel. The p-value is 0.004
The p-value is 0.004 (Round to three decimal places as needed.)

c. What assumptions need to be made to perform this analysis? Select all that apply.
A. The population is normally distributed.
B. The sample mean is large.
C. No assumptions are necessary.
D. The sample is large.
E. The population standard deviation is small.

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

In order to find the critical value, first we need to determine the degrees of freedom given by:

[tex]df=n-1=35-1=34[/tex]

And now since is a right tailed test we need a quantile from the t distribution with 34 degrees of freedom that accumulates 0.05 of the are on the right and 0.95 of the area on the left. We can use the following excel code to find it: "=T.INV(0.95,34)" and we got that [tex]t_{\alpha}=1.691[/tex].

Part b

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{753-744}{\frac{19}{\sqrt{35}}}=2.802[/tex]

And in order to calculate the p value we can do this, since it's a right tailed test:

[tex]p_v = P(t_{34} >2.802)=0.004[/tex]

And the code in excel to find it is given by:"=1-T.DIST(2.802,34,TRUE)"

Conclusion

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject the null hypothesis.

Part c

The 5 assumptions in order to conduct a t test are:

1) The scale of measurement of the data needs to be continuous or ordinal scale

2) We need data from simple random sample sampling, in order to be representative.

3) We need data from a normal distribution, that's called normality assumption.

4) We need a large sample size.

5) And we need homogeneity of variance on the measurements.

Based on this the best options for this case are:

A. The population is normally distributed.

D. The sample is large.