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Topic: **Test of hypothesis for a single sample****Question:**
One of the major measures of the quality of service provided by any organization is the speed with which it responds to customer complaints. A large family held department store selling furniture and flooring had undergone a major expansion in the past several years. In particular, the flooring department had expanded from 2 installation crews to an installation supervisor, measurer, and 15 installation crews. The store had the business objectives of improving its response to complaints. The variable of interest was defined as the number of days between when the complaint was made and when it was resolved. Data were collected from 50 complaints that were made in the last year. It follows:
54 5 35 137 31 27 152 2 123 81 74 27 11 19 126 110 110 29 61 35 94 31 26 5 12 4 165 32 29 28 29 25 26 1 14 13 13 10 5 27 4 52 30 22 36 26 20 23 33 68
a) The installation supervisor claims that the mean number of days between the receipt of a complaint and the resolution of the complaint is 20 days. At 0.05 level of significance, is there evidence that the claim is not true ( ie. that the mean number of days is different from 20)
b) What assumption about the population distribution is needed in order to conduct the t test in (a)
c) Do you think that the assumption needed in order to conduct the t test in (a) is valid? Explain
I know (b) is that you have to assume it's normally distributed.
For (a) do I use the formula t = xbar - population mean / S / root n ??

July 18, 2019 / By Chasity

ANSWER: Conclusion: H1 is true. The average number of days is less than 20. Why???? SINGLE SAMPLE TEST, ONE-TAILED, 6 - Step Procedure for t Distributions, "one-tailed test" Step 1: Determine the hypothesis to be tested. Lower-Tail H0: μ ≥ μ0 H1: μ < μ0 or Upper-Tail H0: μ ≤ μ0 H1: μ > μ0 hypothesis test (lower or upper) = lower Step 2: Determine a planning value for α [level of significance] = 0.05 Step 3: From the sample data determine x-bar, s and n; then compute Standardized Test Statistic: t = (x-bar - μ0)/(s/SQRT(n)) x-bar: Estimate of the Population Mean (statistical mean of the sample) = 43.04 n: number of individuals in the sample = 50 s: sample standard deviation = 41.9 μ0: Population Mean = 20 significant digits = 2 Standardized Test Statistic t = ( 43.04 - 20 )/( 41.9 / SQRT( 50 )) = 3.89 Step 4: Use Students t distribution, 'lookup' the area to the left of t (if lower-tail test) or to the right of t (if upper-tail test) using Students t distribution Table or Excel TDIST(x, n-1 degrees_freedom, 1 tail) =TDIST( 3.89 , 49 , 1 ) Step 5: Area in Step 4 is equal to P value [based on n -1 = 49 df (degrees of freedom)] = 0 Table look-up value shows area under the 49 df curve to the left of t = 3.89 is (approx) probability = 0 Step 6: For P ≥ α, fail to reject H0; and for P < α, reject H0 with 95% confidence. Conclusion: H1 is true Note: level of significance [α] is the maximum level of risk an experimenter is willing to take in making a "reject H0" or "conclude H1" conclusion (i.e. it is the maximum risk in making a Type I error).

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Did you like the answer? We found more questions related to the topic: **Test of hypothesis for a single sample**

The question is not asking for the sample size, it is asking for the **amount you should use a possible cost**. You want a conservative estimate for the worst case scenario, so you are 99% confident that the worst case scenario would be 12,603.8 or the last point within your boundary on your CI with alpha =.01

well you simply just leave the question blank and then drop out of university, then marry earl and live on the roof of montannas. JWANNNNNNNNNA.

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The K-S test looks for any difference in the distributions, the mann-whitney test looks for differences in the ranks. (Or differences in the medians, if you assume that the shape of the distribution is the same in each group).

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