Understanding Basic Statistics
Understanding Basic Statistics
7th Edition
ISBN: 9781305254060
Author: Charles Henry Brase, Corrinne Pellillo Brase
Publisher: Cengage Learning
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Chapter 10.1, Problem 14P

(a)

To determine

The level of significance, null and alternative hypothesis & determine whether we should use a left-tailed, right-tailed, or two-tailed test.

(a)

Expert Solution
Check Mark

Answer to Problem 14P

Solution: The level of significance is α=0.01. The null hypothesis is H0:μd=0 and alternative hypothesis HA:μd0. This is a two-tailed test.

Explanation of Solution

The level of significance is defined as the probability of rejecting the null hypothesis when it is true, it is denoted by α=0.01.

Null hypothesis H0:μd=0

Alternative hypothesis HA:μd0

Since HA:μd0, this is a two-tailed test.

(b)

To determine

To find: The sampling distribution that should be used along with assumptions and compute the value of the sample test statistic.

(b)

Expert Solution
Check Mark

Answer to Problem 14P

Solution: The sampling distribution of d¯ is Student’s t-distribution. The sample test statistic t is 1.17.

Explanation of Solution

Calculation:

  A B d = A - B (dd¯)2
  12.7 9.8 2.9 3.24
  13.4 14.5 -1.1 4.84
  12.8 10.7 2.1 1
  12.1 14.2 -2.1 10.24
  11.6 13 -1.4 6.25
  11.1 12.9 -1.8 8.41
  14.2 10.9 3.3 4.84
  15.1 10 5.1 16
  12.5 14.1 -1.6 7.29
  12.3 13.6 -1.3 5.76
  13.1 9.1 4 8.41
  15.8 10.2 5.6 20.25
  10.3 17.9 -7.6 75.69
  12.7 11.8 0.9 0.04
  11.1 7 4.1 9
  15.7 9.2 6.5 29.16
Sum 206.5 188.9 17.6 210.42
Average 12.90625 11.80625 1.1 13.15125

The d distribution is mound shaped and symmetrical and we have a random sample of n=16 paired differences, so we can assume that sample test statistics follows a Student’s t-distribution with d.f=n1=15.

sd=(dd¯)2n1sd=210.42161sd=3.7454

Using μd=0,n=16,sd=3.7454,d¯=1.1. The test statistic t is

t=(d¯μd)sdnt=(1.10)3.745416t=1.1748t1.17

(c)

To determine

To find: The P-value of the test statistic and sketch the sampling distribution showing the area corresponding to the P-value.

(c)

Expert Solution
Check Mark

Answer to Problem 14P

Solution: The P-value of the test statistic is 0.2602.

Explanation of Solution

Calculation:

We have t = 1.17

Pvalue=P(t<1.17)+P(t>1.17)

Using Table 4 from the Appendix to find the specified area:

0.25<Pvalue<0.50

Graph:

To draw the required graphs using the Minitab, follow the below instructions:

Step 1: Go to the Minitab software.

Step 2: Go to Graph > Probability distribution plot > View probability.

Step 3: Select ‘t’ and enter d.f = 15.

Step 4: Click on the Shaded area >X value.

Step 5: Enter X-value as 1.17 and select ‘Both Tail’.

Step 6: Click on OK.

The obtained distribution graph is:

Understanding Basic Statistics, Chapter 10.1, Problem 14P

Pvalue= 2(0.1301)Pvalue0.2602

(d)

To determine

Whether we reject or fail to reject the null hypothesis and whether the data is statistically significant for a level of significance of 0.01.

(d)

Expert Solution
Check Mark

Answer to Problem 14P

Solution: The P-value >α, hence we fail to reject H0. The data is not statistically significant for a level of significance of 0.01.

Explanation of Solution

The P-value of 0.2602 is greater than the level of significance (α

) of 0.01. Therefore we don’t have enough evidence to reject the null hypothesis H0, hence the data is not statistically significant for a level of significance of 0.01.

(e)

To determine

The interpretation for the conclusion.

(e)

Expert Solution
Check Mark

Answer to Problem 14P

Solution: There is not enough evidence to conclude that there is difference in the population average birthrate and death rate in this region.

Explanation of Solution

The P-value of 0.2602 is greater than the level of significance (α

) of 0.01. Therefore we don’t have enough evidence to reject the null hypothesis H0. There is not enough evidence to conclude that there is difference in the population average birthrate and death rate in this region.

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Chapter 10 Solutions

Understanding Basic Statistics

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