A survey of 64 medical labs revealed that the mean price charged for a certain test was Rs. 120, with a standard deviation of 60. Test whether the data indicates that the mean price of this test is more than Rs. 100 at 5% level of significance. I have solved this question but I don't know whether the answer is correct or not. Solution: H_0: mean = 120 (null hypothesis) H_1: mean > 100 (alternative hypothesis) we will use z test as the sample count is more than 30 z = |120-100|/60/sq.root(64) z = 2.67 at 5% of significance, the critical value of z is 1.96. Since the z value we obtained is more than 1.96, so we reject the null hypothesis and therefore the mean price of the test is more than 100 Please tell whether the answer is correct or there is some mistake in this. Help is appreciated.

Dymnembalmese2n 2022-09-26 Answered
A survey of 64 medical labs revealed that the mean price charged for a certain test was Rs. 120, with a standard deviation of 60. Test whether the data indicates that the mean price of this test is more than Rs. 100 at 5% level of significance.
I have solved this question but I don't know whether the answer is correct or not.
H0: mean = 120 (null hypothesis) H1: mean > 100 (alternative hypothesis)
we will use z test as the sample count is more than 30
z = | 120 100 | / 60 / 64 z = 2.67
at 5% of significance, the critical value of z is 1.96. Since the z value we obtained is more than 1.96, so we reject the null hypothesis and therefore the mean price of the test is more than 100
Please tell whether the answer is correct or there is some mistake in this. Help is appreciated.
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Answers (1)

doraemonjrlf
Answered 2022-09-27 Author has 8 answers
The null Hypothesis is that the mean price is Rs. 100, not Rs. 120. (This may only be a typo on your part.)
Your calculation of the z statistic looks correct (and given the sample size it does make sense to use a z test).
However, since this is a one-tail test, the critical value at 5% significance is only 1.64, not 1.96. Of course this does not change your decision to reject the null.
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