Make 4 claims Goodness of fit claim for the categorical data (web browser) claim

Question

Make 4 claims Goodness of fit claim for the categorical data (web browser) claim for discrete data, (# of social media accounts) claim for continuous data, (minutes of studying) and an Independence claim or Test of Homogeneity for the contingency table) These claims are in sentence form. Test each of your claims using a level of significance of.02 Using the p-value method. Show all steps. The test values can be found by the calculator if possible. The p-values must be found using Excel. State which formula you used. There is a total of 4 Hypothesis Tests (one for each claim).

Explanation / Answer

1.Here we have to test the hypothesis that,

H0 : mu = 0 Vs H1 : mu not= 0

where mu is population mean number of social media account.

Assume alpha = level of significance = 0.02

Here test statistic follows t-distribution.

We can find one sample t test in MINITAB.

steps :

ENTER data into MINITAB sheet --> STAT --> Basic statistics --> 1 sample t --> variables : number of accounts on social media --> test mean : 0 --> optiions --> Confidence level : 0.98 --> Alternative : not equal --> ok --> ok

————— 29-07-2017 05:56:37 ————————————————————

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One-Sample T: no of accounts on socialmedia

Test of mu = 0 vs mu not = 0

Variable N Mean StDev SE Mean
no of accoun 35 3.086 1.380 0.233

Variable 98.0% CI T P
no of accoun ( 2.516, 3.655) 13.23 0.000

Test statistic= 13.23

P-value = 0.000

P-value < alpha

Reject H0 at 0.02 level of significance.

Conclusion : The population mean number of accounts on social media is differ than 0.

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Similarly we have to test,

H0 : mu = 0 Vs H1 : mu not= 0

where mu is population mean number of minutes studying yesterday.

One-Sample T: number of minutes studying yest

Test of mu = 0 vs mu not = 0

Variable N Mean StDev SE Mean
number of mi 35 82.4 85.9 14.5

Variable 98.0% CI T P
number of mi ( 47.0, 117.9) 5.67 0.000

Test statistic = 5.67

P-value =0.000

P-value < alpha

Reject H0 at 2% levelof significance.

Conclusion : The population mean number of minutes studying yesterday is differ than 0.

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Again we have to test the hypothesis that,

H0 : The variables opinion and gender are independent.

H1 : The variables opinion and gender are not independent.

Assume alpha = 0.02

Here test statistic follows X2-distribution.

We can do this test in MINITAB.

steps :

ENTER data into MINITAB sheet --> Stat --> Tables --> Chi square test --> Columns containing the table --> select all the columns --> ok

Chi-Square Test: C1, C2

Skipping rows and/or columns filled with zeros.

Expected counts are printed below observed counts

C1 C2 Total
1 9 8 17
8.26 8.74

2 2 3 5
2.43 2.57

3 0 1 1
0.49 0.51

4 6 6 12
5.83 6.17

Total 17 18 35

Chi-Sq = 0.067 + 0.063 +
0.076 + 0.071 +
0.486 + 0.459 +
0.005 + 0.005 +
DF = 3
* WARNING * 2 cells with expected counts less than 1.0
* Chi-Square approximation probably invalid
4 cells with expected counts less than 5.0

P-value = 0.8783

P-value > alpha

Fail to reject H0 at 2% level of significance.

Conclusion : The variables opinion and gender are independent.

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