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larisa [96]
3 years ago
11

Each year, all final year students take a mathematics exam. It is hypothesised that the population mean score for this test is 1

15. 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.
Mathematics
1 answer:
Yuri [45]3 years ago
7 0

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] .

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