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# Null hypothesis, decision rule, two-tailed test, z value

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6
Marks: 1 Please answer questions 6 - 9 based on the following information.

A firm has an order process time of at least 20 minutes. After a change in the process, a researcher wishes to determine if the processing time for orders in a warehouse averages less than 20 minutes, at a 99% confidence level. The researcher uses a sample of 80 orders from a monthly run of 10,000 orders and determines a sample mean of 19 minutes with a standard deviation of 3 minutes.
The null hypothesis is:

a. H1 : &#956; = 20 minutes
b. H0 : &#956; = 20 minutes
c. H0 : &#956; >20 minutes
d. H0 : &#956; <20 minutes
..

7
Marks: 1 This is an example of a:
a. an upper-tailed test.
b. a lower-tailed test.
c. a two-tailed test.
..

8
Marks: 1 The decision rule is:
a. Fail to reject H0 if z is between -2.58 and +2.58; reject otherwise.
b. Fail to reject H0 if z is between -1.96 and +1.96; reject otherwise.
c. Fail to reject H0 if z is > +2.33; reject otherwise.
d. Fail to reject H0 if z is < -2.33; reject otherwise.
e. Fail to reject H0 if z is > -2.33; reject otherwise.
..

9
Marks: 1 The researcher should reject the null hypothesis.
a. True
b. False
..

10
A sample of 60 observations yields a mean of 160 and a sample standard deviation of 40. A sample of 50 observations of another population yields a mean of 149 and a sample standard deviation of 38. A test at the &#945; = 0.05 level of significance is performed for
H0 : µ 1&#8804; µ2 ; H1 : µ1 > µ2 .

This is an example of a(n):
a. lower-tailed test.
b. upper-tailed test.
c. two-tailed test.
..

11
Marks: 1 The value of the test statistic and p-value are, respectively,:
a. 1.48, 0.0694
b. 1.61, 0.027
c. 1.96, 0.035
d. 2.28, 0.537
..

12
Marks: 1 The results indicate that the null hypothesis should:
a. be rejected.
b. not be rejected.
..

13
Marks: 1 Please answer questions 13 and 14 using the following information.
A report indicates that readership of a home improvement magazine is split evenly among men and women. A random sample of 600 subscribers identifies that 330 are women. The results are tested at the ? = 0.05 level of significance that the report is false.

The null hypothesis and computed value for this problem are, respectively,:
a. H0 : &#956; <0.50; z = 0.46
b. H0 : &#956; = 0.50; z = 1.96
c. H0 : &#960; = 0.50; z = 2.45
d. H0 : &#956; &#8800; 300; z = 2.58
..

14
Marks: 1 We should conclude that the report is false.
a. True.
b. False
..

15
Marks: 1 Please answer questions 15 and 16 using the following information.
In a sample of its readership, 144 of 300 readers in one income bracket approved of the editorial positions of a newspaper. In a different income bracket, 224 of 400 readers approved of the editorial positions. The newspaper wishes to determine if there is a significant difference between the approval rates at an &#945; = 0.05 level of significance.

The computed value of z is:
a. 1.645
b. 1.96
c. 2.02
d. 2.1
e. 55.55
..

16
Marks: 1 The test results indicate that there is a significant difference between the approval rates of the separate income brackets.
a. True
b. False
..

17
Marks: 1 The student t distribution:
a. is preferred over the normal distribution for very large size samples.
b. tends to coincide with the normal distribution for large sample sizes.
c. resembles a curve that is heavily skewed to the left.
d. both a and c.
e. none of the above.
..

18
Marks: 1 The student t distribution will tend to lead to less Type I errors for small samples.
a. True
b. False
..

19
Marks: 1 For a lower-tailed test, when the sample mean is 50, the null hypothesis mean is 52.5, s = 7 and n = 25; the student t statistic is:
a. 2.091
b. -1.79
c. -2.5
d. 1.63
..

20
Marks: 1 For sample A, n = 12 and s2 = 240; for sample B, n = 18 and s2 = 360. The pooled variance of the samples is:
a. 20
b. 300
c. 312.86
d. 350.76
e. not calculable without additional data.
..

21
Marks: 1 Please answer questions 21 and 22 based on the following information.
Procedure 1:n = 6; &#963;= 10; s2 = 8;
Procedure 2:n = 6; &#963; = 14; s2 = 12;

The number of degrees of freedom for the t statistic is:
a. 5
b. 10
c. 11
d. 12
..

22
Marks: 1 The pooled variance and t statistic are, respectively :
a. 10; -2.19
b. 12; -0.8
c. 12.5; 1.2
d. 15; 1.0
..

https://brainmass.com/statistics/hypothesis-testing/null-hypothesis-decision-rule-two-tailed-test-value-307356

#### Solution Preview

The answers and the explanations are in highlighted blue. I attached both doc and docx files, both of which have the same answers.

6
Marks: 1 Please answer questions 6 - 9 based on the following information.
A firm has an order process time of at least 20 minutes. After a change in the process, a researcher wishes to determine if the processing time for orders in a warehouse averages less than 20 minutes, at a 99% confidence level. The researcher uses a sample of 80 orders from a monthly run of 10,000 orders and determines a sample mean of 19 minutes with a standard deviation of 3 minutes.
The null hypothesis is:

a. H1 : μ = 20 minutes
b. H0 : μ = 20 minutes
c. H0 : μ >20 minutes
d. H0 : μ <20 minutes
..

7
Marks: 1 This is an example of a:
a. an upper-tailed test.
b. a lower-tailed test.
c. a two-tailed test.
Explanation: This is because H1 : μ <20 minutes

8
Marks: 1 The decision rule is:
a. Fail to reject H0 if z is between -2.58 and +2.58; reject otherwise.
b. Fail to reject H0 if z is between -1.96 and +1.96; reject otherwise.
c. Fail to reject H0 if z is > +2.33; reject otherwise.
d. Fail to reject H0 if z is < -2.33; reject otherwise.
e. Fail to reject H0 if z is > -2.33; reject otherwise.
..
Explanation:
The rejection region of H0 with 99% confidence level is
Z< - 2.33.
9
Marks: 1 The researcher should reject the null hypothesis.