Recall that “very satisfied” customers give the XYZ-Box video game system a rati
ID: 2956541 • Letter: R
Question
Recall that “very satisfied” customers give the XYZ-Box video game system a rating that is at least 42. Suppose that the manufacturer of the XYZ-Box wishes to use the ... of 65 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42a. Letting µ represent the mean composite satisfaction rating for the XYZ-Box, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence supporting the claim that µ exceeds 42. b. In the context of this situation,interpret making a Type I error; interpret making a Type II error. Recall that “very satisfied” customers give the XYZ-Box video game system a rating that is at least 42. Suppose that the manufacturer of the XYZ-Box wishes to use the ... of 65 satisfaction ratings to provide evidence supporting the claim that the mean composite satisfaction rating for the XYZ-Box exceeds 42
a. Letting µ represent the mean composite satisfaction rating for the XYZ-Box, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence supporting the claim that µ exceeds 42. b. In the context of this situation,interpret making a Type I error; interpret making a Type II error.
Explanation / Answer
Find the mean M and the standard deviation S from the 65 ratings. Our null hypothesis is that the true mean exceeds 42.
If ratings are whole numbers anything more than 42 would be at least 43. I would prefer to split the difference and test whether the mean is more than 42.5--but would do the compariaon to 42 if that is really what is required here.
The null hypothesis is that the mean is less than or equal to 42--or 42.5 if I am allowed to do thew test the way I prefer which is a slightly more stringent test for the company. The alternative hypothesis is that the mean is greater than 42 or as I would prefer 42.5.
For my test I will actually consider the situation where in practice the null hypothesis is a mean of exactly 42 or 42.5 and I will reject it only if the deviation is great enough in the direction of higher ratings.
I will compute my mean M and standard deviation S the sample, I will estimate the standard deviation of the sample mean by dividing S by the square root of n-1 or in this case sqrt(64) which is 8, I will get s=S/8 for my standard deviation of sample means.
Now I look at M - 42 or if I am permitted to do so, M - 42.5, Dividing by s I will get x--my standard deviations away from 42 or 42.5.
A negative x cooresponds to a mean less than 42 or 42.5, and I do not reject the null hypothesis. Otherwise I would compare x to the value I find from a table corresponding to my desired degree of certainty on a one sided test. I would use actual F(x) values for this purpse. For 0.01 uncertainty level I look for 0.01 or 0,99. Preferably, I would use a student t-test for 65 degrees of freedom, but using a normal distribution would no be too bad with a sample size this large.
A Type 1 error would mean accepting the claim that the mean is greater than 42 or 42.5 when that is not true. We affirm that the average rating is very satisfied when that is not the case.
A Type 2 error would mean that the average rating is very satisified but we did not draw that conclusion. Thus an average rating in our sample of 45 might not be enough for us to conclude the true average is more then 42 at say a 0.01 certainty level. We would not reject the null hypothesis with confidence--but we would certainly not conclude that the null hypothesis is true, A type 2 error involves a failure to draw a correct conclusion lacking strong enough evidence. It is not entirely clear that this should even be called an error.
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