Question

In: Statistics and Probability

1.For single-sample statistical tests, if the population variance is unknown you should conduct a (ANSWER) test....

1.For single-sample statistical tests, if the population variance is unknown you should conduct a (ANSWER) test.

2.The critical value for a two-tailed one sample z- test is +/-

3.If the z–score for a sample mean is 1.75, and it was a two-tailed z-test, we (ANSWER) the null hypothesis.

4.A researcher reports the results of a single-sample t-test as t(29) = 1.80. The numerical degrees of freedom for this t-test are (ANSWER), which means there were (ANSWER) total participants in the researcher's sample.

5. Based on the information in question #4 above and knowing it is a two-tailed test, the numerical critical t value is: (ANSWER)

Given a population standard deviation of 14, calculate the standard deviation of the mean ( given the following sample sizes (show all your work and round to two decimal places):

6.N=30

7.N=90

If you know the sample mean for worker satisfaction is 77.3, and the estimated standard error of the mean is 10.7, and the sample size is 27 (calculate the df from the sample size), calculate the 95% confidence interval for a non-directional test (show all your work):

8. Confidence Interval (lower and upper limits):

9. Written statement:

Expert Solution

(1)

For single-sample statistical tests, if the population variance is unknown you should conduct a t test.

(2)

The critical value for a two-tailed one sample z- test is +/- 1.96

(3)

If the z–score for a sample mean is 1.75, and it was a two-tailed z-test, we fail to reject the null hypothesis.

(4)

A researcher reports the results of a single-sample t-test as t(29) = 1.80. The numerical degrees of freedom for this t-test are 29, which means there were 30 total participants in the researcher's sample.

(5)

Based on the information in question #4 above and knowing it is a two-tailed test, the numerical critical t value is: 2.0452

(6)

the standard deviation of the mean is given by:

So,

the standard deviation of the mean is 2.5560

(7)

the standard deviation of the mean is given by:

So,

the standard deviation of the mean is 1.4757

(8)

df = 27 - 1 = 26

= 0.05

From Table, critical values of t = 2.0555

Confidence Interval:

77.3 (2.0555 X 10.7)

= 77.3 21.9939

= (55.3061 ,99.2939)

So,

Answer is:

(55.3061 ,99.2939)

(9)

The 95% Confidence interval (55.3061 ,99.2939) is a range of values we are 95% confident will contain the true unknown population mean.

orchestra answered 1 year ago

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