Question

Please answer all the questions

1. High School Graduates Who Take the SAT. The national average for the percentage of high school graduates taking the SAT is 49%, but the state averages vary from a low of 4% to a high of 92%. A random sample of 300 graduating high school seniors was polled across a particular tristate area, and it was found that 195 of them had taken the SAT. Estimate the true proportion of high school graduates in this region who take the SAT with 95% confidence.

2. Salaries for Actuaries Nationwide. The average salary of actuaries who achieve the rank of Fellow is $150,000. An insurance executive wants to see how this compares with Fellows within his company. He checks the salaries of eight Fellows and finds the average salary to be $155,500 with a standard deviation of $15,000. Can he conclude that Fellows in his company make more than the national average, using  = 0.05?

3. A study was conducted by the Department of Zoology at the Virginia Tech to estimate the difference in the amounts of the chemical orthophosphorus measured at two different stations on the James River. Orthophosphorus was measured in milligrams per liter. Fifteen samples were collected from station 1, and 12 samples were obtained from station 2. The 15 samples from station 1 had an average orthophosphorus content of 3.84 milligrams per liter and a standard deviation of 3.07 milligrams per liter, while the 12 samples from station 2 had an average content of 1.49 milligrams per liter and a standard deviation of 0.80 milligram per liter. Find a 95% confidence interval for the difference in the true average orthophosphorus contents at these two stations, assuming that the observations came from normal populations with different variances.

Answer 1

We need to construct the 95% confidence interval for the population proportion.

We have been provided with the following information about the number of favorable cases:

Favorable Cases X =	195
Sample Size N =	300

The sample proportion is computed as follows, based on the sample size N = 300 and the number of favorable cases X = 195:

The critical value for α=0.05 is

The corresponding confidence interval is computed as shown below:

Therefore, based on the data provided, the 95% confidence interval for the population proportion is 0.596<p<0.704, which indicates that we are 95% confident that the true population proportion pp is contained by the interval (0.596, 0.704)

The provided sample mean is Xˉ=155500

and the sample standard deviation is s = 15000,

and the sample size is n = 8

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 150000

Ha: μ > 150000

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05,

and the critical value for a right-tailed test is t_c = 1.895

The rejection region for this right-tailed test is

R=t:t>1.895

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t=1.037≤tc​=1.895,

it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p = 0.1671,

and since p = 0.1671≥0.05,

it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ is greater than 150000, at the 0.05 significance level.

We need to construct the 95% confidence interval for the difference between the population means μ1​−μ2​, for the case that the population standard deviations are not known.

The following information has been provided about each of the samples:

Sample Mean 1 (Xˉ1​) =	3.84
Sample Standard Deviation 1 (s1​) =	3.07
Sample Size 1 (N1​) =	15
Sample Mean 2 (Xˉ2​) =	1.49
Sample Standard Deviation 2 (s2​) =	0.80
Sample Size 2 (N_2) =	12

Based on the information provided, we assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows:

Now, we finally compute the confidence interval:

Therefore, based on the data provided, the 95% confidence interval for the difference between the population means μ1​−μ2​ is 0.605<μ1​−μ2​<4.095, which indicates that we are 95% confident that the true difference between population means is contained by the interval (0.605, 4.095)

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