Question

An advertisement for a word-processing class claims that students who complete the class will, on average, be able to type 60 words per minute (wpm) with a standard deviation of 6 wpm. At the end of the class, 49 students are tested and their mean score is 58.5 wpm.
A. Is this evidence that the true mean is different from 60 wpm? Give a complete answer, using a significance level of .05. (12 points)
B. Suppose the person conducting the research had believed, before collecting data, that the graduates weren't as good as claimed. Would this belief have changed the analysis you performed in part (A)? (12 points)
C. Construct a 99% confidence interval for the true mean of the population. What conclusion can you draw from this interval, and do you have evidence to reject the claim that the average graduate can type 60 wpm? (6 points)

Answer 1

Answer:

Reject at 5%, accept at 1% the null hypothesis

Step-by-step explanation:

Set up hypotheses as

$H_0: \bar x = 60\\H_a: \bar x < 60$

(Left tailed test)

Population std dev = 6

Sample std error = $\frac{6}{\sqrt{49} } \\=0.8555$

Mean difference = -1.5

Since sigma is known we can use Z test

Z = mean diff/std error = -1.7533

p value = 0.039

a) Since p value <0.05 we reject H0.  There is evidence  that the true mean is different from 60 wpm

b) Yes, because his sample would have been biased since he may want to prove his belief so slow or inefficient persons he would have selected in the sample.

c) For 99% confidence interval critical value = 2.58

Confidence interval for population mean = 58.5±2.58*std error

=(56.2928, 60.7072)

Since this contains 60, the hypothesized mean, we accept null hypothesis.

we do not have evidence to reject the claim that the average graduate can type 60 wpm at 1% level of significance.