Fundaments: Statistics 10.6 Use critical values to test the null hypothesis H 0: µ 1 - µ 2 = 20 versus the alternative hypothesis H a : µ 1 - µ 2 =? 20 by setting a equal to .10, .05, .01, and .001....

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Fundaments: Statistics
10.6


Use critical values to test the null hypothesis
H0:
µ1 -
µ2 = 20 versus the alternative hypothesis
H
a:
µ1 -
µ2 =? 20 by setting
a
equal to .10, .05, .01, and .001. How much evidence is there that the difference between
µ1 and
µ2 is not equal to 20?
Missing formula:



10.8


An article in
Fortune
magazine reported on the rapid rise of fees and expenses charged by mutual funds. Assuming that stock fund expenses and municipal bond fund expenses are each approximately normally distributed, suppose a random sample of 12 stock funds gives a mean annual expense of 1.63 percent with a standard deviation of .31 percent, and an independent random sample of 12 municipal bond funds gives a mean annual expense of 0.89 percent with a standard deviation of .23 percent. Let
µ1 be the mean annual expense for stock funds, and let
µ2 be the mean annual expense for municipal bond funds. Do parts
a,
b, and
c
by using the equal variances procedure. Then repeat
a,
b, and
c
using the unequal variances procedure. Compare your results.

  • a) Set up the null and alternative hypotheses needed to attempt to establish that the mean annual expense for stock funds is larger than the mean annual expense for municipal bond funds. Test these hypotheses at the .05 level of significance. What do you conclude? ?

  • b) Set up the null and alternative hypotheses needed to attempt to establish that the mean annual expense for stock funds exceeds the mean annual expense for municipal bond funds by more than .5 percent. Test these hypotheses at the .05 level of significance. What do you conclude? ?

  • c) Calculate a 95 percent confidence interval for the difference between the mean annual expenses for stock funds and municipal bond funds. Can we be 95 percent confident that the mean annual expense for stock funds exceeds that for municipal bond funds by more than .5 percent? Explain. ?


10.18


In the book
Essentials of Marketing Research,
William R. Dillon, Thomas J. Madden, and Neil H. Firtle (1993) present preexposure and postexposure attitude scores from an advertising study involving 10 respondents. The data for the experiment are given in Table 10.3. Assuming that the differences between pairs of postexposure and preexposure scores are normally distributed:

  1. Set up the null and alternative hypotheses needed to attempt to establish that the advertisement increases the mean attitude score (that is, that the mean postexposure attitude score is higher than the mean preexposure attitude score). ?

  2. Test the hypotheses you set up in part
    a
    at the .10, .05, .01, and .001 levels of significance. How much evidence is there that the advertisement increases the mean attitude score? ?

  3. Estimate the minimum difference between the mean postexposure attitude score and the mean preexposure attitude score. Justify your answer. ?


10.34


Suppose two independent random samples of sizesn1 = 9 and
n2 = 7 that have been taken from two normally distributed populations having variances and give sample variances of = 100 and = 20.
1. Test
H0: = versus
H
a: =? with
a
= .05. What do you conclude? 2. Test
H0: = versus
H
a: > with
a
= .05. What do you conclude?
Answered Same DayDec 26, 2021

Answer To: Fundaments: Statistics 10.6 Use critical values to test the null hypothesis H 0: µ 1 - µ 2 = 20...

Robert answered on Dec 26 2021
119 Votes
1
Question 10.6
Two tailed hypothesis test
Null hypothesis H0:
Alternative hypothesis Ha:
,

̅ ̅̅ ̅̅
Degree of freedom
Where

( ) ( )

Now,
Value of t stat

( )
√ (



)

would be used because the applicable test is two tailed test.
2
Here, null hypothesis would be rejected because value of t statistics is much higher than
critical value .
Now,
Degree of freedom = 12
Conclusion
0.10 0.05 3.388 1.782 Null hypothesis would be
rejected.
0.05 0.025 3.388 2.179 Null hypothesis would be
rejected.
0.01 0.005 3.388 3.055 Null hypothesis would be
rejected.
0.001 0.0005 3.388 4.318 Null hypothesis would not be
rejected.
Finally, it can be concluded that strong evidence available to cite that difference between
would not equal to 20.
Question 10.8
(a)
Null hypothesis H0:
Alternative hypothesis Ha:

( ) ( ) ( ) ( )

3
Value of t stat

( )
√ (



)

Degree of freedom = 22
Here, t = 6.64 > 1.717 and 5% level of significance. Therefore, null hypothesis would be
rejected.
(b)
Null hypothesis H0:
Alternative hypothesis Ha:
Value of t stat

( ...
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