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Salaries of entry-level computer engineers have Normal distribution with unknown mean and variance. Than sample standard devi

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Answer #1

i am using minitab to solve the problem.

steps :-

copy the data in minitab and name the column x 1596051930854_blob.png stat 1596051930978_blob.png basic statistics 1596051930897_blob.png 1-sample T 1596051930881_blob.png select summarized data from the drop down menu 1596051930866_blob.png type 3 in sample size, 50000 in sample mean,20 in standard deviation  1596051931005_blob.pngtick perform hypothesized test 1596051930882_blob.pngin hypothesized mean type 80000 1596051930906_blob.pngoptions 1596051930917_blob.png in confidence level type 90 1596051930883_blob.png select alternative hypothesis as mean < hypothesized mean 1596051931066_blob.pngok1596051931005_blob.pngok.

***SOLUTION***

here, we will do 1 sample t test for mean as the population variance is unknown and sample size(n) = 3<30.

hypothesis:-

HOJ = 80000

H, :μ< 80000[ claim ]

the test statistic (t) = -2598.08

p value = 0.000

decision:-

p value = 0.000< 0.10(alpha)

we reject the null hypothesis. There is enough evidence that the average salary of all entry level computer engineer is less than 80000.

*** if you have any doubt regarding the problem ,please write it in the comment box...if satisfied,please UPVOTE.

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