Question 1171911: A random sample of 15 students majoring in computer science has an average SAT score of ๐ฅฬ
= 1173 with a standard deviation of ๐ = 85. Let x be a random variable representing the SAT score for all computer science majors. Assume the distribution of x is mound shaped and symmetric. Previous studies indicate that the average SAT score for computer science major was about ฮผ = 1143. We want to determine if the data indicate that the average SAT score for computer science major should be higher than 1143 using a level of significance of ฮฑ = 10%. Explain why we can use a Studentโs t distribution.
What are the null and alternate hypotheses? Compute the t value of the sample test statistic. Truncate to two decimal places. Estimate the P-value for the test. Do we reject or fail to reject H0? Interpret the results.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Given information:
n = 15 = sample size
xbar = 1173 = sample mean test score
mu = 1143 = population mean test score
s = 85 = sample standard deviation of the test scores
alpha = 0.10 = significance level
Since we don't know sigma (population standard deviation) and because n = 15 is not greater than 30, we must use a student's T distribution. We can refer to it in short as "T distribution".
We can use a T distribution because "the distribution of x is mound shaped and symmetric". This seems to imply that the data is approximately normally distributed (which is often the case with these types of problems).
The degrees of freedom (df) are df = n-1 = 15-1 = 14.
The hypotheses are:
H0: mu = 1143
H1: mu > 1143
where H0 is the null and H1 is the alternative. The claim "average SAT score for computer science major should be higher than 1143" is in the alternative. The alternative having a greater than sign means we have a right tailed test.
Computing the test statistic:
t = (xbar - mu)/(s/sqrt(n))
t = (1173-1143)/(85/sqrt(15))
t = 1.36693529866144
t = 1.37
We round to two decimal places as instructed.
The t test statistic is roughly t = 1.37
To calculate the P-value, we need to use a calculator. Doing so by hand would take way too long. Use the tcdf function on your TI83/TI84 calculator.
If you don't have such a calculator, then here's a free resource
https://stattrek.com/online-calculator/t-distribution.aspx
Make sure "t score" is selected (after "Random Variable")
Type 14 in the degrees of freedom box
Type 1.37 in the t score box
Leave the last box blank
Then hit "calculate" and the answer will show up in that box we left blank. You should get 0.9039
This means P(T < 1.37) = 0.9039
Since we're doing a right tailed test, this means
P(T > 1.37) = 1 - P(T < 1.37)
P(T > 1.37) = 1 - 0.9039
P(T > 1.37) = 0.0961
which is the P-value we're after
P-value = 0.0961
Comparing the p-value to alpha = 0.10, we see that the p-value is smaller than alpha. So we reject the null.
Therefore, we conclude that mu > 1143
Interpretation: The mean SAT score for computer science majors is greater than 1143
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