According to a recent report​ (based on historical​ data), 47% of college student internships are unpaid. A recent survey of 100 college interns at a local university found that 59 had unpaid internships.   Follow the​ five-step ​p-value approach to hypothesis testing​ (see Chapter 9 Powerpoint slides​ #53 - 59 or​ e-text p. 339​ - 340. Use a 0.10 level of significance to determine whether the proportion of college interns that had unpaid internships is different from 0.47. Part 1

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According to a recent report​ (based on historical​ data), 47% of college student internships are
unpaid. A recent survey of 100 college interns at a local university found that 59 had unpaid internships.
 
Follow the​ five-step ​p-value approach to hypothesis testing​ (see Chapter 9 Powerpoint slides​ #53 - 59 or​ e-text p. 339​ - 340. Use a 0.10 level of significance to determine whether the proportion of college interns that had unpaid internships is different from 0.47.
Part 1
Let π be the population proportion. Determine the null​ hypothesis, H0​, and the alternative​ hypothesis, H1.
​(Type integers or decimals. Do not​ round.)
A. H0 : μ = 0.47; H1 : μ ≠ 0.47 ​, because μ is always used in hypothesis testing
B. H0 : π = 0.47; H1 : π ≠ 0.47, because the two hypotheses always use a known or likely to be true value for the population proportion
C. H0 : π = 0.59; H1 : π ≠ 0.59​, since we are trying to prove the sample proportion of 59100= 0.59
is really true
D. H0 : π ≠ 0.47; H1 : π = 0.47​, since H0must never have an equal symbol
 
What is the test​ statistic? ​(Round to two decimal places to the right of the decimal point as​ needed.)
A. ZSTAT =(0.47 − 0.59)(0.59)•(1−0.59)100 = −2.44, using formula​ (9.3)
B. tSTAT =(0.59 − 0.47)0.47100 = +2.55, using formula​ (9.2) since the sample standard deviation is not known
C. ZSTAT =(0.59 − 0.47)(0.47)•(1−0.47)100 = +2.40, using formula​ (9.3)
D. ZSTAT =(0.59 − 0.47)(0.59)•(1−0.47)100 = +2.15​, using formula​ (9.3)
 
What is the​ p-value? ​(Round to three decimal places to the right of the decimal point as​ needed.)
A. ​p-value =​ NORM.S.DIST(2.40,1) = 0.992
B. ​p-value =​ 2*(1-NORM.S.DIST(2.40,1) = 0.016
C. ​p-value =​ 2*NORM.S.DIST(2.40,1) = 1.984
D. ​p-value =​ 1-NORM.S.DIST(2.40,1)
 
What is the final​ conclusion?
A. Do not reject the null hypothesis. There is sufficient evidence that the proportion of college interns that had unpaid interships is different from 0.47 because the​ p-value is less than level of significance.
B. Do not reject the null hypothesis. There is not enough evidence that the proportion of college interns that had unpaid interships is different from 0.47 because the​ p-value is greater than level of significance.
C. Reject the null hypothesis. There is sufficient evidence that the proportion of college interns that had unpaid interships is different from 0.47 because the​ p-value is less than level of significance.
D. Reject the null hypothesis. There is enough evidence to show that the proportion of college interns that had unpaid interships is different from 0.47 because the​ p-value is greater than level of significance.
 
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