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# Testing of hypothesis for proportion

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1-Testing for categories with different proportions. Here are the observed frequencies from four categories; 5,10,10,20. Assume that we want to use a 0.05 significance level to test the claim that the four categories have proportions of 0.20, 0.25, 0.25and 0.30, repectively
a. What is the null hypothesis?
b. What are the expected frequencies for the four categories?
c. What is the value of the test statistic?
d. What is the critical value?
e. What do you conclude about the given claim?

2-No smoking,
The accompanying table summarizes successes and failure when subjects used different methods in trying to stop smoking. The determination of smoking or not smoking was made five month after the treatment was begun, and the data are based on results from the center of disease control and prevention. Use the TI-83/84Plus result {see below} with a 0.05significance level to test the claim that success is independent of the method used. If someone wants to stop smoking, does the choice of the method make a difference?

nicotine gum Nicotine patch
smoking 191 263
non smoking 59 57

TI-83/84 Plus
X-Test
X(X)=2.900233793
p=.0885667054
df=1

3-Solar energy in different weather.
A student of the author lives in home with solar electric system. At the same time each day ,she collected voltage readings from a meter connected to the system and analysis of variance was used with readings obtained on three different types of day: sunny, cloudy, and rainy. The TI-83/84 plus the calculator results are in the margin. Use a 0.05 significance level to test the claim that the mean voltage reading is the same for the three different of day. Is there sufficient evidence to support a claim of different population means? we might expect that a solar system would provide more electric energy on sunny days than on cloudy or rainy days. Can we conclude that sunny days result in greater amounts of electric energy?

TI-83/84
One -way ANOVA
F=38. 03789731
P=1. 3340195e-6
Factor
df=15
SS=1.36333333
↓ MS=3.45722222

One -way ANOVA
 Ms=3.4572222
Error
df=2
SS=6. 91444444
Sxp=.301477841

4-use the excel display which results from the scores listed in table below the sample data are SAT scores on the verbal and math portions of SAT-I and are based on reported statistics from the College Board

verbal
female 646 539 348 623 478 429 298 782 626 533
male 562 525 512 576 570 480 571 555 519 596

math
female 484 489 436 396 545 504 574 352 365 350
male 547 678 464 651 645 673 624 624 328 548

ANOVA
source of variation SS df MS F P-Value F crit
Sample 52635.03 1 52635.03 5.029517 0.031169 4.113161
Columns 6027.025 1 6027.025 0.57591 0.45286 4.113161
Interaction 31528.22 1 31528.22 3.012666 0.09117 4.113161
Within 376748.1 36 10465.23

Total 466938.4 39

part one: Interaction effect - test the null hypothesis that SAT scores are not affected by an interaction between gender and test (verbal/math). What do you conclude?

Part 2: Effect of Type of SAT Test- Assume that SAT scores are not affected by an interaction between gender and the type of test (verbal/Math). Is there sufficient evidence to support the claim that gender has an effect on SAT scores?

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#### Solution Preview

1-Testing for categories with different proportions. Here are the observed frequencies from four categories; 5,10,10,20. Assume that we want to use a 0.05 significance level to test the claim that the four categories have proportions of 0.20, 0.25, 0.25and 0.30, respectively
a. What is the null hypothesis?
b. What are the expected frequencies for the four categories?
c. What is the value of the test statistic?
d. What is the critical value?
e. What do you conclude about the given claim?

Category O prop E (O-E)2/E
1 5 0.2 9 1.777778
2 10 0.25 11.25 0.138889
3 10 0.25 11.25 0.138889
4 20 0.3 13.5 3.12963
Total 45 1 45 5.185185

a). What is the null hypothesis?
H0: The four categories have proportions of 0.20, 0.25, 0.25and 0.30, respectively.

b). What are the expected frequencies for the four categories?

Category O prop E
1 5 0.2 9
2 10 0.25 11.25
3 10 0.25 11.25
4 20 0.3 13.5
Total 45 1 45

c). What is the value of the test statistic?
=5.185185
d). What is the critical value?
Critical value = =3.84
e). What do you conclude about the given claim?
Reject the null hypothesis as the value of test statistic is greater than the critical value. Thus the proportions of the four categories are different from 0.20, 0.25, 0.25and 0.30, respectively.

2-No smoking,
The accompanying table summarizes successes and failure when subjects used different methods in trying to stop smoking. The determination of smoking or not smoking was made five month after the treatment was begun, and the data are based on results from the center of disease control and prevention. Use the TI-83/84Plus result {see below} with a 0.05significance level to test the claim that success is independent of the method used. If someone ...

#### Solution Summary

Step by step method for testing the hypothesis under 5 step approach is discussed here. Excel template for each problem is also included. This template can be used to obtain the answers of similar problems.

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