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Verizon [17]
3 years ago
10

the number off standard deviations between a score and the mean score indicated by a z-score. what was Cindy's z-score on the sc

ience test
Mathematics
1 answer:
aalyn [17]3 years ago
5 0
Z score of 0 gives you the mean for the problem;  The mean μ is 70 Now you can use this value of the mean in the definition of z score for the second case and find the standard deviation. Z score is the number of standard deviations away from the mean.  Therefore σ = (1)(74-70) = 4.
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Using the z-distribution, we have that:

a)

The hypothesis test is:

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The p-value is of 0.0668.

The critical value is z^{\ast} = -1.28

b) Since the <u>test statistic is less than the critical value</u> for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

Item a:

At the null hypothesis, it is <u>tested if the mean is of 7.4 minutes</u>, that is:

H_0: \mu = 7.4

At the alternative hypothesis, it is <u>tested if the mean is lower than 7.4 minutes</u>, that is:

H_1: \mu < 7.4

We have the <u>standard deviation for the population</u>, hence, the z-distribution is used.

The test statistic is given by:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • \sigma is the standard deviation of the sample.
  • n is the sample size.

For this problem, the values of the <em>parameters </em>are: \overline{x} = 7.1, \mu = 7.4, \sigma = 1.2, n = 36.

Hence, the value of the <em>test statistic</em> is:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{7.1 - 7.4}{\frac{1.2}{\sqrt{36}}}

z = -1.5

Using a z-distribution calculator, the p-value is of 0.0668.

Also using a calculator, considering a <u>left-tailed test</u>, as we are testing if the mean is less than a value, the critical value with a <u>significance level of 0.1</u> is of z^{\ast} = -1.28.

Item b:

Since the <u>test statistic is less than the critical value</u> for the left-tailed test, there is enough evidence to conclude that the null hypothesis will be rejected.

You can learn more about the use of the z-distribution to test an hypothesis at brainly.com/question/16313918

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