Question 1107647: 39. A Statistics teacher claims that 60% of his students will pass the class with an ”A”. Due to change of text book and educational objectives there is a worry that the number of students getting an ”A” will be smaller now. A random sample of 80 students shows that 43 got an ”A”. Test the teachers claim at significance level of 5%.
(a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?
. (b) Identify the sampling distribution you will us: the standard normal or Student’s t. Explain the rationale for your choice. What is the value of the sample test statistic?
. (c) Find (or estimate) the P − value. Sketch the sampling distribution and show the area corre- sponding to the P − value.
(d) Find the critical value(s).
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! (a) level of significance is 0.05
null hypothesis: x = 0.60
alternative hypothesis x not = 0.60
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since the alternative hypothesis has "not =", we have a two tailed test
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the sample size is > 30, so we can use a normal distribution
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the population mean proportion is 0.60
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the sample mean proportion is 43/80 = 0.5375
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the sample standard deviation is square root(0.60 * (1-0.60) / 80) = 0.0548
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(b) z-score(test statistic) = (0.5375 - 0.60) / 0.0548 = -1.14
the p-value associated with this z-score is 0.1271
since this is a two-tailed test, the p-value is 0.1271 + 0.1271 = 0.2542
since 0.2542 > 0.05, we accept the null hypothesis
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critical value is found for 0.05/2 = 0.025 is 1.960
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