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I THINK I DID THIS CORRECT, BUT FEEL STUCK ON #16 #17 IF IT IS WORDED CORRECTLY?

ID: 3226409 • Letter: I

Question

I THINK I DID THIS CORRECT, BUT FEEL STUCK ON #16 #17 IF IT IS WORDED CORRECTLY??

For this final project you will analyze the variables in the data set “perio sex al.xls”. You are interested in examining whether periodontal attachment loss (AL) varies by sex. You have measured all variables among participants entering an emergency room. The following variables were collected as part of an intake interview.

Part C)

11)

What are the null and the alternative hypotheses ?

The null is: The mean periodontal attachment loss varies by sex.

Ho: =

The alternative hypotheses is: The mean periodontal attachment does not vary by sex.

Ha:   

12)

What is the dependent variable?

Periodontal attachment loss. It is a continuous variable.

13)

What is the independent variable? What type of variable is it?

Sex. It is a binary variable.

14)

What type of statistical test would you perform?

I choose to do a two-sample t-test because the standard deviation was not known on the sex variable and the population mean is unknown. The data does not violate the assumption for a t-test.

15)

Descriptive statistics (Quantitative data):

Statistic

al

age

sex

Minimum

Maximum

Mean

Std. deviation

Nbr. of observations

558

558

558

0.000

1.000

0.606

0.489

Minimum

0.3

26

0

0.300

6.300

3.312

1.328

Maximum

6.3

93

1

1st Quartile

2.6

60

0

Median

3.3

69

1

3rd Quartile

3.988

75

1

Mean

3.312

67.344

0.606

Variance (n-1)

1.763

145.188

0.239

Standard deviation (n-1)

1.328

12.049

0.489

                    

t (Observed value)

-45.175

|t| (Critical value)

1.962

DF

1114

p-value (Two-tailed)

< 0.0001

alpha

0.05

Test interpretation:

H0: The difference between the means is equal to 0.

Ha: The difference between the means is different from 0.

As the computed p-value is lower than the significance level alpha=0.05, one should reject the null hypothesis H0, and accept the alternative hypothesis Ha.

The risk to reject the null hypothesis H0 while it is true is lower than 0.01%.

16) Interpret your finding

I choose to reject the null hypothesis that states the mean periodontal attachment loss varies by sex based on the p-value = 0.0001 and that value is less than .05 level of significance. There is statistical significance that periodontal attachment levels do not vary by sex.

17)Is there Statistical Significance?

Yes, because the 95% confidence interval means that 95% of the intervals would contain the mean of the sample. It has a .95 probability of containing the mean of the sample. The p-value of .0001 is less than .05

Descriptive statistics (Quantitative data):

Statistic

al

age

sex

Minimum

Maximum

Mean

Std. deviation

Nbr. of observations

558

558

558

0.000

1.000

0.606

0.489

Minimum

0.3

26

0

0.300

6.300

3.312

1.328

Maximum

6.3

93

1

1st Quartile

2.6

60

0

Median

3.3

69

1

3rd Quartile

3.988

75

1

Mean

3.312

67.344

0.606

Variance (n-1)

1.763

145.188

0.239

Standard deviation (n-1)

1.328

12.049

0.489

                    

t (Observed value)

-45.175

|t| (Critical value)

1.962

DF

1114

p-value (Two-tailed)

< 0.0001

alpha

0.05

Test interpretation:

H0: The difference between the means is equal to 0.

Ha: The difference between the means is different from 0.

As the computed p-value is lower than the significance level alpha=0.05, one should reject the null hypothesis H0, and accept the alternative hypothesis Ha.

The risk to reject the null hypothesis H0 while it is true is lower than 0.01%.

Explanation / Answer

16) Interpret your finding

I choose to reject the null hypothesis that states the mean periodontal attachment loss varies by sex based on the p-value = 0.0001 and that value is less than .05 level of significance. There is statistical significance that periodontal attachment levels do not vary by sex.

17)Is there Statistical Significance?

Yes, because the 95% confidence interval means that 95% of the intervals would contain the mean of the sample. It has a .95 probability of containing the mean of the sample. The p-value of .0001 is less than .05

(Your answer is correct)