PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Chapter 10.2, Problem 55E

(a)

To determine

To explain why the sample results give evidence for the alternative hypothesis.

(a)

Expert Solution
Check Mark

Explanation of Solution

It is given that:

  x¯1=6.37x¯2=5.91n1=92n2=86s1=0.60s2=0.93

The given claim that: difference in the means.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  H0:μ1=μ2Ha:μ1μ2

Where we have,

  μ1= the true mean reliability rating of all Angle bank customers.

  μ2= the true mean reliability rating of all Hispanic bank customers.

We then know that the sample means 6.37 and 5.91 differ which agrees with the alternative hypothesis Ha:μ1μ2 that the means are different and thus the sample results give evidence for the alternative hypothesis.

(b)

To determine

To calculate the standardized test statistics and P -value.

(b)

Expert Solution
Check Mark

Answer to Problem 55E

The P -value is P<0.001 or P=0.0002 and the t=3.892 .

Explanation of Solution

It is given that:

  x¯1=6.37x¯2=5.91n1=92n2=86s1=0.60s2=0.93

The given claim that: difference in the means.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  H0:μ1=μ2Ha:μ1μ2

Where we have,

  μ1= the true mean reliability rating of all Angle bank customers.

  μ2= the true mean reliability rating of all Hispanic bank customers.

Now, find the test statistics:

  t=(x¯1x¯2)(μ1μ2)s12n1+s22n2=6.375.9100.60292+0.93286=3.892

Now, the degree of freedom will be:

  df=min(n11,n21)=min(921,861)=85

Since student’s T distribution table in the appendix does not contains the value of df=85 so we will take the nearest value df=80 . So the P -value will be:

  P<2(0.0005)=0.001

On the other hand by using the calculator command: 2×tcdf(3.892,1E99,85) which results in the P -values as 0.0002 .

Thus, the P -value is P<0.001 or P=0.0002 and the t=3.892 .

(c)

To determine

To explain what conclusion would you make.

(c)

Expert Solution
Check Mark

Explanation of Solution

From part (a) and (b), we have,

  x¯1=6.37x¯2=5.91n1=92n2=86s1=0.60s2=0.93

The appropriate hypotheses for this is:

  H0:μ1=μ2Ha:μ1μ2

And the P -value is P<0.001 or P=0.0002 and the t=3.892 .

And we know that if the P -value is less than or equal to the significance level then the null hypothesis is rejected, then,

  P<0.05Reject H0

Thus, we conclude that there is convincing evidence of a difference in the mean reliability ratings of all Angle and Hispanic bank customers.

Chapter 10 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 10.1 - Prob. 11ECh. 10.1 - Prob. 12ECh. 10.1 - Prob. 13ECh. 10.1 - Prob. 14ECh. 10.1 - Prob. 15ECh. 10.1 - Prob. 16ECh. 10.1 - Prob. 17ECh. 10.1 - Prob. 18ECh. 10.1 - Prob. 19ECh. 10.1 - Prob. 20ECh. 10.1 - Prob. 21ECh. 10.1 - Prob. 22ECh. 10.1 - Prob. 23ECh. 10.1 - Prob. 24ECh. 10.1 - Prob. 25ECh. 10.1 - Prob. 26ECh. 10.1 - Prob. 27ECh. 10.1 - Prob. 28ECh. 10.1 - Prob. 29ECh. 10.1 - Prob. 30ECh. 10.1 - Prob. 31ECh. 10.1 - Prob. 32ECh. 10.1 - Prob. 33ECh. 10.1 - Prob. 34ECh. 10.1 - Prob. 35ECh. 10.1 - Prob. 36ECh. 10.2 - Prob. 37ECh. 10.2 - Prob. 38ECh. 10.2 - Prob. 39ECh. 10.2 - Prob. 40ECh. 10.2 - Prob. 41ECh. 10.2 - Prob. 42ECh. 10.2 - Prob. 43ECh. 10.2 - Prob. 44ECh. 10.2 - Prob. 45ECh. 10.2 - Prob. 46ECh. 10.2 - Prob. 47ECh. 10.2 - Prob. 48ECh. 10.2 - Prob. 49ECh. 10.2 - Prob. 50ECh. 10.2 - Prob. 51ECh. 10.2 - Prob. 52ECh. 10.2 - Prob. 53ECh. 10.2 - Prob. 54ECh. 10.2 - Prob. 55ECh. 10.2 - Prob. 56ECh. 10.2 - Prob. 57ECh. 10.2 - Prob. 58ECh. 10.2 - Prob. 59ECh. 10.2 - Prob. 60ECh. 10.2 - Prob. 61ECh. 10.2 - Prob. 62ECh. 10.2 - Prob. 63ECh. 10.2 - Prob. 64ECh. 10.2 - Prob. 65ECh. 10.2 - Prob. 66ECh. 10.2 - Prob. 67ECh. 10.2 - Prob. 68ECh. 10.2 - Prob. 69ECh. 10.2 - Prob. 70ECh. 10.2 - Prob. 71ECh. 10.2 - Prob. 72ECh. 10.2 - Prob. 73ECh. 10.2 - Prob. 74ECh. 10.3 - Prob. 75ECh. 10.3 - Prob. 76ECh. 10.3 - Prob. 77ECh. 10.3 - Prob. 78ECh. 10.3 - Prob. 79ECh. 10.3 - Prob. 80ECh. 10.3 - Prob. 81ECh. 10.3 - Prob. 82ECh. 10.3 - Prob. 83ECh. 10.3 - Prob. 84ECh. 10.3 - Prob. 85ECh. 10.3 - Prob. 86ECh. 10.3 - Prob. 87ECh. 10.3 - Prob. 88ECh. 10.3 - Prob. 89ECh. 10.3 - Prob. 90ECh. 10.3 - Prob. 91ECh. 10.3 - Prob. 92ECh. 10.3 - Prob. 93ECh. 10.3 - Prob. 94ECh. 10.3 - Prob. 95ECh. 10.3 - Prob. 96ECh. 10.3 - Prob. 97ECh. 10.3 - Prob. 98ECh. 10.3 - Prob. 99ECh. 10.3 - Prob. 100ECh. 10.3 - Prob. 101ECh. 10.3 - Prob. 102ECh. 10 - Prob. R10.1RECh. 10 - Prob. R10.2RECh. 10 - Prob. R10.3RECh. 10 - Prob. R10.4RECh. 10 - Prob. R10.5RECh. 10 - Prob. R10.6RECh. 10 - Prob. R10.7RECh. 10 - Prob. T10.1SPTCh. 10 - Prob. T10.2SPTCh. 10 - Prob. T10.3SPTCh. 10 - Prob. T10.4SPTCh. 10 - Prob. T10.5SPTCh. 10 - Prob. T10.6SPTCh. 10 - Prob. T10.7SPTCh. 10 - Prob. T10.8SPTCh. 10 - Prob. T10.9SPTCh. 10 - Prob. T10.10SPTCh. 10 - Prob. T10.11SPTCh. 10 - Prob. T10.12SPTCh. 10 - Prob. T10.13SPTCh. 10 - Prob. AP3.1CPTCh. 10 - Prob. AP3.2CPTCh. 10 - Prob. AP3.3CPTCh. 10 - Prob. AP3.4CPTCh. 10 - Prob. AP3.5CPTCh. 10 - Prob. AP3.6CPTCh. 10 - Prob. AP3.7CPTCh. 10 - Prob. AP3.8CPTCh. 10 - Prob. AP3.9CPTCh. 10 - Prob. AP3.10CPTCh. 10 - Prob. AP3.11CPTCh. 10 - Prob. AP3.12CPTCh. 10 - Prob. AP3.13CPTCh. 10 - Prob. AP3.14CPTCh. 10 - Prob. AP3.15CPTCh. 10 - Prob. AP3.16CPTCh. 10 - Prob. AP3.17CPTCh. 10 - Prob. AP3.18CPTCh. 10 - Prob. AP3.19CPTCh. 10 - Prob. AP3.20CPTCh. 10 - Prob. AP3.21CPTCh. 10 - Prob. AP3.22CPTCh. 10 - Prob. AP3.23CPTCh. 10 - Prob. AP3.24CPTCh. 10 - Prob. AP3.25CPTCh. 10 - Prob. AP3.26CPTCh. 10 - Prob. AP3.27CPTCh. 10 - Prob. AP3.28CPTCh. 10 - Prob. AP3.29CPTCh. 10 - Prob. AP3.30CPTCh. 10 - Prob. AP3.31CPTCh. 10 - Prob. AP3.32CPTCh. 10 - Prob. AP3.33CPTCh. 10 - Prob. AP3.34CPTCh. 10 - Prob. AP3.35CPT
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