Question

Each year, all final year students take a mathematics exam. It is hypothesised that the population mean score for this test is 115. It is known that the population standard deviation of test scores is 17. A random sample of 25 students take the exam. The mean score for this group is 101. a)Calculate the 90% confidence interval for the population mean test score. Give your answers to 2 decimal places.

Answer 1

Answer:

90% confidence interval for the population mean test score is [95.40 , 106.59]

Step-by-step explanation:

We are given that the population mean score for mathematics test is 115. It is known that the population standard deviation of test scores is 17.

Also, a random sample of 25 students take the exam. The mean score for this group is 101.

The, pivotal quantity for 90% confidence interval for the population mean test score is given by;

        P.Q. = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean = 101

              $\sigma$ = population standard deviation

              n = sample size = 25

So, 90% confidence interval for the population mean test score, $\mu$ is ;

P(-1.6449 < N(0,1) < 1.6449) = 0.90

P(-1.6449 < $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.6449) = 0.90

P(-1.6449 * $\frac{\sigma}{\sqrt{n} }$ < ${Xbar-\mu}$ < 1.6449 * $\frac{\sigma}{\sqrt{n} }$ ) = 0.90

P(X bar - 1.6449 * $\frac{\sigma}{\sqrt{n} }$ < $\mu$ < X bar + 1.6449 * $\frac{\sigma}{\sqrt{n} }$ ) = 0.90

90% confidence interval for $\mu$ = [ X bar - 1.6449 * $\frac{\sigma}{\sqrt{n} }$ , X bar + 1.6449 * $\frac{\sigma}{\sqrt{n} }$ ]

                                                  = [ 101 - 1.6449 * $\frac{17}{\sqrt{25} }$ , 10 + 1.6449 * $\frac{17}{\sqrt{25} }$ ]

                                                  = [95.40 , 106.59]

Therefore, 90% confidence interval for the population mean test score is [95.40 , 106.59] .