Answer to Question #348922 in Statistics and Probability for maya

Question #348922

1. A researcher estimates that the average height of the buildings of 30 or more

stories in a large city is at least 700 feet. A random sample of 10 buildings is

selected, and the heights in feet are shown. At  = 0.025, is there enough

evidence to reject the claim?

485 511 841 725 615

520 535 635 616 582


1
Expert's answer
2022-06-08T16:16:05-0400
"\\bar{x}=\\dfrac{1}{10}(485 +511+ 841+ 725+ 615""+520 +535 +635 +616+ 582)=606.5""s^2=\\dfrac{\\sum_i(x_i-\\bar{x})^2}{n-1}=\\dfrac{1}{10-1}((485-606.5)^2""+(511-606.5)^2+(841-606.5)^2""+(725-606.5)^2+(615-606.5)^2""+(520-606.5)^2+(535-606.5)^2""+(635-606.5)^2+(616-606.5)^2""+(582-606.5)^2)=\\dfrac{107084.5}{9}""s=\\sqrt{s^2}=\\sqrt{\\dfrac{107084.5}{9}}\\approx109.08"

The parameter is average content of fruit concentrate per bottle.

The following null and alternative hypotheses need to be tested:

"H_0:\\mu\\ge700"

"H_1:\\mu<700"

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is "\\alpha = 0.025," "df=n-1=9" and the critical value for a left-tailed test is "t_c =-2.262156."

The rejection region for this left-tailed test is "R = \\{t:t<-2.262156\\}."

The t-statistic is computed as follows:



"t=\\dfrac{\\bar{x}-\\mu}{s\/\\sqrt{n}}=\\dfrac{606.5-700}{109.08\/\\sqrt{10}}=-2.7106"


Since it is observed that "t=-2.7106<-2.262156=t_c," it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, "df=9" degrees of freedom, "t=-2.7106" is "p=0.011987," and since "p=0.011987<0.025=\\alpha," it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean "\\mu"

is less than 700, at the "\\alpha = 0.025" significance level.


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