5,60,60,61,61,62,62,62,63,63,63,64,64,64,64,64,64,64,65,65,65,65,66,67,67,67,67,68,68,68,70,70,72,72,72,72,74,74,510 Mean = 2954/39 = 75.7436 Standard deviation = 72.1299   Confidence Level is 0.90 Or 90% then x=0.10 T value at x=0.10 dt=(39-1=38) is =1.69 90% confidence interval for μ= x̅ ± t x s/√n =75.7436± 1.69x 72.1299/√39 =(55.657, 94.603)   We are 90% confident that true mean lies between 55.657 to 94.603

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5,60,60,61,61,62,62,62,63,63,63,64,64,64,64,64,64,64,65,65,65,65,66,67,67,67,67,68,68,68,70,70,72,72,72,72,74,74,510



Mean = 2954/39 = 75.7436

Standard deviation = 72.1299

 

Confidence Level is 0.90

Or 90% then x=0.10

T value at x=0.10 dt=(39-1=38) is =1.69

90% confidence interval for μ= ± t x s/√n

=75.7436± 1.69x 72.1299/√39

=(55.657, 94.603)

 

We are 90% confident that true mean lies between 55.657 to 94.603



When confidence level is 95%

x=0.05

T value =0.05 =38 is 2.02

95% confidence interval= 75.743±2.02x 72.1299/√39

=(51.748, 98.512

We are 95% confident that true mean lies between 51.748 to 98.512



We confidence level is 99% 

x=0.01 tc=2.71

Then 99% confidence interval is = ± t x s/√n

=75.1299 +- 2.71x 72.1299/  √39

=(43.811, 106.448)

We are 99% confident that true population mean lies between 43.811 and 106.448

 

Sample mean= 75.1299

Sample standard deviation= 72.1299

Sample size= 39



Hypothesis

Sample mean x̅= 75.744

Sample standard deviation s= 72.13

Sample size n= 39

 

The following null and alternative hypotheses need to be tested 

 

Ho μ= 72

Ha μ≠ 72

It's a two-tailed test which will be a t-test for one mean. With unknown population standard deviation.

 

Significance level =0.05

Critical value= tc= 2.024

 

t-test = x̅-μo / (s/√n) = 75.744-72/(72.13/√39) =0.324

 

|t|=0.324 ≤tc =2.024 

Null hypothesis is not rejected 

 

P= 0.7476 

Since p=0.7476≥0.05 concludes that the null hypothesis is not rejected. 

 

Conclusion, null hypothesis is not rejected, there is not enough evidence to claim the population mean is different than 72, at 0.05 significance level. 

 

Confidence interval 

95% confidence interval is 52.362<μ<99.126

Question: Write a minimum of a five-sentence summary of your findings. Use the following questions to guide you in writing this summary.
(1) What conclusions can you make based on the results you collected? 
(2) Do you think there were any bias that might have made your outcome what it was?
(3) Based on the data sampling we collected at the beginning of the semester, do you believe it was a good representation of your conclusion?
(4) After completing this assignment/project, are there any areas of statistics that are still not clear to you? Be as specific as possible. 

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