Take this test as you would take a test in class. For this test, do the followin

Question

Take this test as you would take a test in class. For this test, do the following (a) Identify the claim and state Ho and Ha (b) Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed and whether to use a z-test, a t-test, or a chi-square test. Explain your reasoning. Option 1: Find the critical value(s), identify the rejection region(s), and find Option 2: Find the appropriate standardized test statistic and the P-value. (c) Choose one of the options. If convenient, use technology the appropriate standardized test statistic. (d) Decide whether to reject or fail to reject the null hypothesis. (e) Interpret the decision in the context of the original claim n 80% of coffee drinkers think t the taste of a shop's coffee is very important in determining where they purchase their coffee. In a random sample of 36 coffee drinkers, 78% think that the taste of a shop's coffee is very important in determining where they purchase their coffee. At 0.10, is there enough evidence to support the owner's claim? (Adapted from Harris Interaern

Explanation / Answer

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P < 0.80

Alternative hypothesis: P > 0.80

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

zcritical = 1.28

Critical region is z > 1.28

= sqrt[ P * ( 1 - P ) / n ]

= 0.0667

z = (p - P) /

z = - 0.2999

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a one-tailed test, the P-value is the probability that the z-score is more than - 0.2999. We use the Normal Distribution Calculator to find P(z > - 0.2999)

Thus, the P-value = 0.6179

Interpret results. Since the P-value (0.6179) is greater than the significance level (0.10), we have to accept the null hypothesis.

Fail to reject the null hypothesis, There is not enough evidence to support the owner's claim.

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