Question 1204650
find the mean and standad deviation of the sample.
i get meqan = 49.1111 and standard deviation = 2.6914
standard error = standard deviation / sqrt(sample size) = 2.6914 / sqrt(9) = .8971
t-score = (x-m)/s = (49.1111 - 47.5) / .8971 = 1.7959.
area under normal distributuion curve to the right of that t-score with 8 degrees of freedom = .0551
two tailed critical p-factor on each end of 95% confidence interval = .025.
this is higher than that, indicating the resuls are not significant.
there is not enough information to say that the population mean is not what is claimed to be.
i solved this with my ti-84 plus calculator and confirmed the answer through the use of the online statistical calculator at <a href = "https://www.statskingdom.com/130MeanT1.html" target = "_blank">https://www.statskingdom.com/130MeanT1.html</a>
here are the results from that calculator.


<img src = "http://theo.x10hosting.com/2023/110101.jpg">


the results indicae that the p-value is 10.22.
the p-value i got from my ti-84 plus was .0511.
the .0511 p-value i got was for the high end tail of the confidence interval.
that would be alpha/2.
alpha is therefore twice that = 10.22
the critical alpha is .05 on both tails.
that's .025 for each tail.
when comparing p-values, you would either compare .0511 to .025 or .1022 to .05.
either way, the test p-value is higher than the critical p-value by a fairly wide margin.
the results from the online calculator tell you that the 95% confidence interval is 47.0977 to 51.1245.
49.1111 is within tht interval, indicating the results are not significant at .05 two tailed significance level.