Question

We want to estimate the mean starting salary in a population of graduates. Approximately normally distributed. Standard deviation = 10 In a random sample of 31 graduates, the mean starting salary was 61.8 and the standard deviation was 12. 1. What is the lower (smaller) endpoint of the 90% confidence interval for the population mean? 2. What is the upper (larger) endpoint of the 90% confidence interval for the population mean? 3. If null hypothesis = 60 and alternate hypothesis is different from 60, test the null hypothesis that the population mean is 60 against the alternative hypothesis that the population mean is different from 60, at the 10% significance level. What is the conclusion? A. FAIL TO REJECT H. (accept Ho) B. REJECT Ho in favor of HA 4. If we wanted the length of the confidence interval to be 4 or less, what is the minimum sample size that would be necessary?

Answer 1

The data is given as

Sample size(n)=31

Sample mean()=61.8

T

Sample SD(s)=12

1. and 2.

The 90% confidence interval for population mean is given by

lī * n n 10.05,n-1)

$i.e.\;(61.8\pm \frac{12}{\sqrt{31}}*t_{0.05,30})$

12 * 1.697261 i.e. (61.83 /31

i.e. (61.8 + 3.658044894)

The lower endpoint of the 90% confidence interval is 58.14195511

The upper endpoint of the 90% confidence interval is 65.45804489

3.

Let the population mean be denoted by $\mu$ . Here we are to test

$H_0:\mu=60\;\;against\;\;H_1:\mu\neq 60$

As the value 60 lies within the 90% confidence interval, it can be concluded that the population mean is not different from 60 at 10% level of significance.

4.

For the length of the confidence interval to be 4, we should have the margin of error to be 2

Let the required sample size be n₀

$\therefore \frac{10}{\sqrt{n_0}}*\tau_{0.05}=2$

$\therefore \frac{10}{\sqrt{n_0}}*1.645=2$

$\therefore \frac{10}{2}*1.645=\sqrt{n_0}$

$\therefore \sqrt{n_0}=8.225$

$\therefore \sqrt{n_0}=67.650625\;i.e.\; 68$

The required sample size is 68

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