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musickatia [10]
3 years ago
14

Calculate the data value that corresponds to each of the following z-scores.

Mathematics
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

a) 90.1

b) $35.6

c) 3.0915 hours

Step-by-step explanation:

The z-score measures how many standard deviations a score X is above or below the mean.

It is given by the following formula:

Z = \frac{X - \mu}{\sigma}

In which \mu is the mean and \sigma is the standard deviaition.

In all three cases, we have to find X

a. Final exam scores: Allison’s z-score = 2.30, μ = 74, σ = 7.

Z = \frac{X - \mu}{\sigma}

2.30 = \frac{X - 74}{7}

X - 74 = 7*2.3

X = 90.1

b. Weekly grocery bill: James’ z-score = –1.45, μ = $53, σ = $12.

Z = \frac{X - \mu}{\sigma}

-1.45 = \frac{X - 53}{12}

X - 53 = -1.45*12

X = 35.6

Mean and standard deviation in dollars, so the answer also in dollars.

c. Daily video game play time: Eric’s z-score = –0.79, μ = 4.00 hours, σ = 1.15 hours.

Z = \frac{X - \mu}{\sigma}

-0.79 = \frac{X - 4}{1.15}

X - 4 = -0.79*1.15

X = 3.0915

Mean and standard deviation in hours, answer in hours.

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Answer:

a. z=3.09. Yes, it can be concluded that the population mean is greater than 50.

b. z=1.24. No, it can not be concluded that the population mean is greater than 50.

c. z=2.22. Yes, it can be concluded that the population mean is greater than 50.

Step-by-step explanation:

We have a hypothesis test for the mean, with the hypothesis:

H_0: \mu\leq50\\\\H_a:\mu> 50

The sample size is n=55 and the population standard deviation is 6.

The significance level is 0.05.

We can calculate the standard error as:

\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{55}}=0.809

For a significance level of 0.05, the critical value for z is zc=1.644. If the test statistic is bigger than 1.644, the null hypothesis is rejected.

a. If the sample mean is M=52.5, the test statistic is:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{52.5-50}{0.809}=\dfrac{2.5}{0.809}=3.09

The null hypothesis is rejected, as z>zc and falls in the rejection region.

b. If the sample mean is M=51, the test statistic is:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51-50}{0.809}=\dfrac{1}{0.809}=1.24

The null hypothesis failed to be rejected, as z<zc and falls in the acceptance region.

c. If the sample mean is M=51.8, the test statistic is:

z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51.8-50}{0.809}=\dfrac{1.8}{0.809}=2.22

The null hypothesis is rejected, as z>zc and falls in the rejection region.

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