Question

A sample of students from an introductory psychology class were polled regarding the number of hours they spent in studying for the last exam. All students anonymously submitted the number of hours on a 3 by 5 card. There were 24 individuals in the one section of the course polled. There data are below: 4.5, 22, 7, 14.5, 9, 9, 3.5, 8, 11, 7.5, 18, 20, 7.5, 9, 10.5, 15, 19, 2.5, 5, 9, 8.5, 14, 20, 8.

A. based on the sample results, find the 95% confidence interval.

B. Interpret the result

C.Do you expect a 90% confidence interval to be wider or narrower and why.

Answer 1

Here in this Question the sample of students from introductory psychology class were polled regarding the no of hours they spend for studying for last exam.

Here the sample of size 24 students spend the no. of hour is studying is given below,

4.5, 22, 7, 14.5, 9, 9, 3.5, 8, 11, 7.5, 18, 20, 7.5, 9, 10.5, 15, 19, 2.5, 5, 9, 8.5, 14, 20, 8.

Now here, Based on sample results we have to compute,

A) The 95% Confidence Interval for mean time spent by students in studying.

Here the population standard deviation is unknown and the sample size is less than n< 30 so we have to Use T score.

The 95% confidence interval is,

N: 24

Sum, Σx: 262

Mean, x̄: 10.916

Variance, s2: 31.3405

Steps

s2 =Σ(xi - x̄)2/N - 1

=(4.5 - 10.916666666667)2 + ... + (8. - 10.916666666667)2/24 - 1

=720.83333333333/23

= 31.340579710145

s = √31.340579710145

s = 5.5982657770192

The provided sample mean is \bar X = 10.9166Xˉ=10.9166 and the sample standard deviation is s = 5.5982s=5.5982. The size of the sample is n = 24n=24 and the required confidence level is 95%.

Based on the provided information, the critical t-value for \alpha = 0.05α=0.05 and df = 23df=23 degrees of freedom is t_c = 2.069tc​=2.069.

The 95% confidence for the population mean \muμ is computed using the following expression

Therefore, based on the information provided, the 95 % confidence for the population mean

CI=(10.9166−2.069×5.5982/√24,10.9166+2.069×5.5982​/√24)

= (10.9166 - 2.364, 10.9166 + 2.364)

=(10.9166−2.364,10.9166+2.364)

= (8.553, 13.281)

B) Interpretation :

From the above confidence interval we conclude that we 95% confidence that the true mean time in hours that students spend in studying for last exam is in bet 8.5 to 13.2 Hours.

C) The 95% confidence interval is the way of finding the mean value, now here the 90% Confidence Interval is shorter or narrower confidence interval than 95% because in 90% Confidence Interval we 90% Confident that the mean is lie in bet bounds.

Here is the simple answer of your Question.

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