Homework Solution – Hypothesis Testing Homework Solution – Hypothesis Testing


See some sample statistics solutions

Sample Question 1: Give two reasons why the research hypothesis is not tested directly.

Solution: Because the mechanics of a hypothesis test are based upon determining how unlikely or extreme the obtained sample results are, under the assumption that the null hypothesis is true. Therefore, the way hypothesis testing is structured is that we reject Ho (and accept Ha) only if there is enough evidence to contradict Ho.

Sample Question 2: How should a projected hypothesis test be modified if you’re particularly concerned about

(a) the type I error?

(b) the type II error?

Solution: (a) Reduce alpha

(b) Increase alpha.

Sample Question 2: Consult the power curves in Figure 11.7 to estimate the approximate detection rate, rounded to the nearest tenth, for each of the following situations:

(a) a four-point effect, with a sample size of 13.

(b) a ten-point effect, with a sample size of 29.

(c) a seven-point effect with a sample size of 18. (Interpolate)

Solution: (a) Approximately 0.23

(b) Approximately 0.94

(c) Approximately 0.60

Hypothesis testing example

Sample Question 3: In Review Question 11.12 on page 263, instead of testing a hypothesis, you might prefer to construct a confidence interval for the mean weight of all 2-pound boxes of candy during a recent production shift.

(a) Given a population standard deviation of .30 ounce and a sample mean weight of 33.09 ounces for a random sample of 36 candy boxes, construct a 95 percent confidence interval.

(b) Interpret this interval, given the manufacturer’s desire to produce boxes of candy that on the average exceed 32 ounces.

Solution: (a) The population standard deviation is available to, so the normal distribution can be used. Therefore, we obtain that the 95% confidence interval is given

\[CI=\left( \bar{X}-{{z}_{\alpha /2}}\times \frac{\sigma}{\sqrt{n}},\,\,\bar{X}+{{z}_{\alpha /2}}\times \frac{\sigma}{\sqrt{n}} \right)\]

where \({{z}_{\alpha /2}}\) corresponds to the two-tailed z-critical value for \(\alpha =0.05\). Therefore, we find that

\[CI=\left( {33.09}-{1.96}\times \frac{0.30}{\sqrt{36}},\,\,{33.09}+{1.96}\times \frac{0.30}{\sqrt{36}} \right)=\left( {32.992},\,\,{33.188} \right)\]

(b) The interpretation is that we are 95% confident that the actual population mean \(\mu\) is contained by the interval \(\left( {32.992},\,\text{ }{33.188} \right)\).

Sample Question 4: It’s tempting to claim that once a particular 95 percent confidence interval has been constructed, it includes the unknown population characteristic with a probability of .95. What is wrong with this claim?

Solution: Such claim implies that \(\mu\) is a random variable, when in fact \(\mu \) is a fixed, constant, population parameter.

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