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Salsk061 [2.6K]
2 years ago
9

Suppose that a random sample of 10 adults has a mean score of 62 on a standardized personality test, with a standard deviation o

f 4 . (A higher score indicates a more personable participant.) If we assume that scores on this test are normally distributed, find a 90% confidence interval for the mean score of all takers of this test. Then find the lower limit and upper limit of the 90% confidence interval.
Mathematics
1 answer:
aniked [119]2 years ago
5 0

Answer:

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.645*\frac{4}{\sqrt{10}} = 2.08

The lower end of the interval is the sample mean subtracted by M. So it is 62 - 2.08 = 59.92

The upper end of the interval is the sample mean added to M. So it is 62 + 2.08 = 64.08.

The 90% confidence interval for the mean score of all takers of this test is between 59.92 and 64.08. The lower end is 59.92, and the upper end is 64.08.

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