Inferential Statistics

[Name of the Author]

[Name of the institute]

Case 11

1. Hypothesis testing statistical significance for difference between means1

Z-test2

Critical region2

Test Statistics2

Decision2

2. Confidence intervals3

Conclusion4

Case 25

1. For Proportion (New product placement)5

Conclusion5

2. For Proportion (Competitive products)6

Conclusion6

Case 37

1. Hypothesis testing statistical significance for one standard deviation7

-test7

Level of Significance:7

Rejection region:8

Test Statistics:8

Decision8

References9

Case 1

1. Hypothesis testing statistical significance for difference between means

For this test, the following data will be used

n = 50

= 94.4

= 102.4

s=27.57

a (at 95% CI) = 0.05 = 0.05

z (at 95% CI) = 1.96

The population standard deviation in this case is unknown (which is mandatory for conducting z-test). Still, if n>30, the sample standard deviation is assumed to be an unbiased estimator of population standard deviation after application of some adjustment.

Adjusting the sample standard deviation for normalization

Z-test

Null Hypothesis: The difference between the mean town spending and mean national spending is zero.

Alternate Hypothesis: The mean town spending is greater than the mean national spending. Mathematically,

H0:

H1:

Level of Significance

a=0.05

Critical region

zcritical = 1.96

Test Statistics

=== 2.48

Decision

Since the Z statistics (z = 2.48) exceeds the critical value of z critical (Z critical = 1.96) the null hypothesis that “The difference between the mean town spending and mean national spending is zero.” is rejected in favor of the alternate hypothesis that “The mean town spending is greater than the mean national spending.” at 5% level of significance.

Through this hypothesis testing, statistically significant evidence has come out to authenticate that the mean town spending is greater than the mean national spending.

2. Confidence intervals

Confidence interval around the mean amount women in her area spend will be calculated using the following data

n = 50

= 94.4

= 102.4

s=27.57

a (at 95% CI) = 0.05 = 0.05/2 = 0.025

z (at 95% CI) = 1.96

The population standard deviation in this case is unknown (which is mandatory in CI around mean for normal distribution). Still, if n>30, the sample standard deviation ...