Question

2. Suppose 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. The authors plan to administer the test to all third-grade students in New Jersey. Construct a 95% con dence interval for the mean score of all New Jersey third graders. b. Suppose the same test is given to 200 randomly selected third graders from Iowa, producing a sample average of 62 points and sample standard deviation of 11 points. Construct a 90% con dence interval for the di erence in mean scores between Iowa and New Jersey. c. Can you conclude with a high degree of con dence that the population means for Iowa and New Jersey students are di erent

Answer 1

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.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~  $t_n_-_1$

where, $\bar X$ = sample average score = 58 points

             s = sample standard deviation = 8 points

            n = sample of third-grade students = 100

             $\mu$ = 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, $\mu$ is;

P(-1.987 < $t_9_9$ < 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 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.987) = 0.95

P( $-1.987 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.987 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.987 \times {\frac{s}{\sqrt{n} } }$ ]

                                       = [ $58-1.987 \times {\frac{8}{\sqrt{100} } }$ , $58+1.987 \times {\frac{8}{\sqrt{100} } }$ ]

                                       = [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.  =  $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$  ~  $t__n_1_+_n_2_-_2$

where, $s_p = \sqrt{\frac{(n_1-1)\times s_1 + (n_2-1)\times s_2}{n_1+n_2-2} }$

                = $\sqrt{\frac{(200-1)\times 11 + (100-1)\times 8}{200+100-2} }$ = 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, ($\mu_1-\mu_2$) is;

P(-1.645 < $t_2_9_8$ < 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 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < 1.645) = 0.90

90% confidence interval for ($\mu_1-\mu_2$) = [ ${(\bar X_1-\bar X_2) -1.645\times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ , ${(\bar X_1-\bar X_2) +1.645\times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ]

= [ ${(62-58) -1.645\times {3.163 \times \sqrt{\frac{1}{200}+\frac{1}{100} } }$ , ${(62-58) +1.645\times {3.163 \times \sqrt{\frac{1}{200}+\frac{1}{100} } }$ ]

= [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.