Answer :
Answer:
(a) A 95% confidence interval for the mean score of all New Jersey third graders is [56.41, 59.59] .
(b) A 90% confidence interval for the difference in mean scores between Iowa and New Jersey is [3.363, 4.637] .
(c) Yes, we are 90% confident that the population means for Iowa and New Jersey students are different.
Step-by-step explanation:
We are given that a new standardized test is given to 100 randomly selected third-grade students in New Jersey. The sample average score Y on the test is 58 points and the sample standard deviation sY is 8 points.
(a) Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average score = 58 points
s = sample standard deviation = 8 points
n = sample of third-grade students = 100
[tex]\mu[/tex] = population mean score of all New Jersey third graders
Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.
So, a 95% confidence interval for the population mean, [tex]\mu[/tex] is;
P(-1.987 < [tex]t_9_9[/tex] < 1.987) = 0.95 {As the critical value of t at 99 degrees of
freedom are -1.987 & 1.987 with P = 2.5%} P(-1.987 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.987) = 0.95
P( [tex]-1.987 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.987 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-1.987 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X+1.987 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X-1.987 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X+1.987 \times {\frac{s}{\sqrt{n} } }[/tex] ]
= [ [tex]58-1.987 \times {\frac{8}{\sqrt{100} } }[/tex] , [tex]58+1.987 \times {\frac{8}{\sqrt{100} } }[/tex] ]
= [56.41, 59.59]
Therefore, a 95% confidence interval for the mean score of all New Jersey third graders is [56.41, 59.59] .
Now, the same test is given to 200 randomly selected third graders from Iowa, producing a sample average of 62 points and a sample standard deviation of 11 points.
(b) Firstly, the pivotal quantity for finding the confidence interval for the difference in population mean is given by;
P.Q. = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t__n_1_+_n_2_-_2[/tex]
where, [tex]s_p = \sqrt{\frac{(n_1-1)\times s_1 + (n_2-1)\times s_2}{n_1+n_2-2} }[/tex]
= [tex]\sqrt{\frac{(200-1)\times 11 + (100-1)\times 8}{200+100-2} }[/tex] = 3.163
Here for constructing a 90% confidence interval we have used a two-sample t-test statistics because we don't know about population standard deviations.
So, a 90% confidence interval for the difference in two population means, ([tex]\mu_1-\mu_2[/tex]) is;
P(-1.645 < [tex]t_2_9_8[/tex] < 1.645) = 0.90 {As the critical value of t at 298 degrees of
freedom are -1.645 & 1.645 with P = 5%} P(-1.645 < [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < 1.645) = 0.90
90% confidence interval for ([tex]\mu_1-\mu_2[/tex]) = [ [tex]{(\bar X_1-\bar X_2) -1.645\times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] , [tex]{(\bar X_1-\bar X_2) +1.645\times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ]
= [ [tex]{(62-58) -1.645\times {3.163 \times \sqrt{\frac{1}{200}+\frac{1}{100} } }[/tex] , [tex]{(62-58) +1.645\times {3.163 \times \sqrt{\frac{1}{200}+\frac{1}{100} } }[/tex] ]
= [3.363, 4.637]
Therefore, a 90% confidence interval for the difference in mean scores between Iowa and New Jersey is [3.363, 4.637] .
(c) Yes, we are 90% confident that the population means for Iowa and New Jersey students are different because in the above interval 0 is not included.