Athabasca Fishing lodge is located on lake athabasca in northern canada. In one
ID: 3225637 • Letter: A
Question
Athabasca Fishing lodge is located on lake athabasca in northern canada. In one of its recent brochures, the lodge advertises that 75% of its guests catch northern pike over 20 pounds. suppose that last summer 64 out of a random sample of 83 guests did, in fact, catch northern pike weighing over 20 pounds. Does this indicate that the population proportion of guests who catch pike over 20 pounds is different from 75% (either higher or lower)? significance=0.05
a) What is the level of significance? stat eht null and alternative hypotheses.
b) What sampling distribution will you use? Do you think the sample size is sufficiently large? Explain. Compute the value of the sample test statistic.
c) Find the P-value of the test statistic. Sketch the sampling distribution and show the area corresponding to the P-value.
d) Based on your answers in parts (a) and (c), will you reject or fail to reject the null hypothesis? are the data statistically significant at level of significance?
e) interpret your conclusion in the context of the application
Explanation / Answer
a) H0: p0 = 0.75 vs. H1: p0 0.75
b) If the p-value is less than or equal to the significance level , i.e., p-value , then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value > , the significance level then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.
The hypothesis test in this question is: H0: p0 = 0.75 vs. H1: p0 0.75
The test statistic is:
z = ( 0.7710843 - 0.75 ) / ( ( 0.75 * (1 - 0.75 ) / 83 )
z = 0.443607
c) The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.
H1: p p0; p-value is the area in the tails greater than |z|
H1: p < p0; p-value is the area to the left of z
H1: p > p0; p-value is the area to the right of z
phat = r/n = 64/83 = 0.77
z = (phat - p)/SQRT{pq/n} = (0.77 - 0.75)/SQRT{0.75*0.25/83} = 0.4
The p-value = P( Z > |z| )
= P( Z < -0.443607 ) + P( Z > 0.443607 )
= 2 * P( Z < -0.443607 )
= 0.6573268
d) Since the p-value is greater than the significance level of 0.05 we fail to reject the null hypothesis and conclude p = 0.75 is plausible.
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