Conduct a hypothesis test (using either the p-value approach or the critical val

Question

Conduct a hypothesis test (using either the p-value approach or the critical value approach) to determine if the proportion of all business students at Rocky University who were not involved in some type of cheating is less than that of all business students elsewhere. Use = 0.05. Do not forget to include the correctly worded hypotheses. PLEASE HELP ALSO IN THE CORRECT WORDING OF THE HYPOTHESIS.

Here is what I have from the data:

Here is what I have so far, but not sure what to do from here:

Cheating    54%

Not Cheating 46%

H0: P= 0.466

H1: P< 0.466

n=90

# of people who didn't cheat = 10

Sample proportion p = 10/90 = 0.111

z = ((0.111-0.466)/(sqrt(0.466*(1-0.466)/90))

z = -6.751

Student Copied from Internet Copied on Exam Collaborated on Individual Project Gender 1 NO YES NO Male 2 YES NO YES Female 3 NO YES YES Female 4 NO NO NO Male 5 NO YES NO Male 6 NO NO YES Female 7 NO NO YES Male 8 YES YES YES Male 9 YES NO NO Female 10 NO NO YES Female 11 NO YES NO Male 12 YES YES NO Female 13 YES NO YES Male 14 YES NO NO Female 15 NO NO YES Female 16 YES YES NO Female 17 YES YES YES Male 18 NO YES NO Female 19 YES NO YES Male 20 NO NO YES Male 21 NO YES NO Female 22 NO NO NO Male 23 YES NO NO Male 24 YES YES YES Female 25 NO NO YES Male 26 NO YES NO Male 27 NO YES YES Female 28 YES NO YES Female 29 YES NO NO Female 30 YES YES NO Male 31 NO NO NO Male 32 NO NO YES Female 33 NO YES YES Female 34 NO NO YES Male 35 YES YES NO Female 36 NO YES YES Female 37 NO YES NO Female 38 NO NO YES Male 39 YES NO NO Male 40 NO YES YES Male 41 NO NO NO Female 42 YES YES YES Male 43 YES YES NO Female 44 YES NO NO Female 45 NO NO NO Female 46 YES YES YES Female 47 NO NO YES Male 48 YES NO NO Male 49 NO NO NO Female 50 YES NO YES Male 51 NO NO YES Female 52 YES NO YES Male 53 YES YES NO Male 54 NO NO YES Female 55 NO NO NO Male 56 NO YES NO Male 57 NO NO YES Male 58 YES NO NO Female 59 YES YES NO Female 60 YES NO NO Male 61 NO NO YES Male 62 YES YES NO Female 63 NO NO NO Female 64 YES YES YES Male 65 YES YES YES Female 66 YES YES NO Male 67 YES NO YES Female 68 NO NO YES Female 69 YES YES YES Male 70 YES NO YES Male 71 NO NO YES Female 72 YES NO NO Female 73 YES NO YES Female 74 YES YES NO Female 75 YES NO NO Female 76 YES YES YES Female 77 YES YES NO Female 78 NO NO NO Male 79 NO YES NO Female 80 NO NO NO Male 81 YES NO YES Female 82 YES YES NO Male 83 YES NO YES Male 84 YES YES YES Female 85 YES YES NO Male 86 YES YES NO Female 87 NO NO YES Male 88 YES YES YES Female 89 YES YES YES Male 90 YES YES YES Female

Explanation / Answer

Solution:-

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

Null hypothesis: P = 0.46

Alternative hypothesis: P < 0.46

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 0.05. 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).

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

= 0.05254

z = (p - P) /

z = - 6.64

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 less than - 6.64. We use the Normal Distribution Calculator to find P(z < - 6.64) = 0.04.

Thus, the P-value = less than 0.00001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

From this we can conclude that we have sufficient evidence in the favor of the claim that proportion of all business students at Rocky University who were not involved in some type of cheating is less than that of all business students elsewhere.

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