The soccer team is conducting a fundraiser selling long-sleeved t-shirts for $14 and short-sleeved t-shirts for $10. So far the
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Answer:
We conclude that the mean lead concentration for all such medicines is less than 17 mu.
Step-by-step explanation:
We are given below are the lead concentrations in mu;
6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20
We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.
<u><em>Let </em></u><u><em> = mean lead concentration for all such medicines.</em></u>
So, Null Hypothesis, :
= 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}
Alternate Hypothesis, :
< 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}
The test statistics that would be used here <u>One-sample t test</u> <u>statistics</u> as we don't know about population standard deviation;
T.S. = ~
where, = sample mean lead concentration =
= 12.25 mu
s = sample standard deviation = = 6.96 mu
n = sample size = 10
So, <u><em>test statistics</em></u> = ~
= -2.16
The value of t test statistics is -2.16.
<u>Now, the P-value of the test statistics is given by the following formula;</u>
P-value = P( < -2.16) = <u>0.031</u>
<u><em>Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test.</em></u><em> Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis</u></em><em>.</em>
Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.