Two Statistics Problems
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1. On a college bulletin, it is stated that the weights of the male students are normally distributed with mean = 165lb and a standard deviation = 16lb. A research student decided to verify the claim of the school by taking a sample of 64 students. The average weight of the sampled student is 170lb. Test the hypothesis:
Ho: mean = 165lb
Ha: mean does not equal 165lb
Using (1) 5% level of significance
(2) 1% level of significance
2. Suppose the temperature during June is normally distributed with mean = 20°C and standard deviation = 3.33 deg. A is the event that temperature is between 21.11°C and 26.66°C. B is the event that temperature is over 25°C. Event C is that temperature is below freezing point. Find the following probabilities:
(a) P(A)
(b) P(A and B)
(c) P(C)
(d) P(C given B)
(e) P(neither A nor B)
(f) Find the expected number of days in June when event B will occur.
Table for Hypothesis Testing
Level of Significance= 0.10, 0.05, 0.01,0.005, 0.002
Critical Values of z for One-Tailed Tests -1.28 or 1.28 -1.645 or 1.645 -2.33 or 2.33 -2.58 or 2.58 -2.88 or 2.88
Critical Values of z for Two-Tailed Tests -1.645 and 1.645 -1.96 and 1.96 -2.58 and 2.58 -2.81 and 2.81 -3.08 and 3.08
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Solution Summary
This solution involves determining the significance of hypotheses and finding probabilities.
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Question 1
In order to test this hypothesis, we need to do the following. First of all, we have to find to find a confidence interval around the mean proposed by the null hypothesis. The confidence interval is built with respect to the significance level, so let's start with the case in which the required significance level is 5%. The confidence interval will then be a set of values in which the sample mean could fall with 95% probability if it were true that the population mean is 165.
After this we will be able to check if we can reject the null hypothesis. For example, let's say that we find this confidence interval to be [162, 168]. This will mean that, given a population mean of 165, random samples of 64 students will yield a sample mean that will be in that interval with 95% probability. Therefore, if the actual mean we observe (170 in this case) is outside that interval, we can reject the null hypothesis, because if the population mean were truly 165, then the sample mean should have fallen inside that interval almost surely (95% prob). Let's see how to find the confidence interval.
Let's call Xbar to the sample mean and s to the to find values 'a' and 'b' such that:
Prob ( Xbar > 165 + a ) = 0.025
Prob (Xbar < 165 - b) = 0.025
[that is, the probability that we're outside the interval [165-b, 165+a] is 0.05]
Now, given that the population is normally distributed, the distribution of the sample mean (Xbar) will also be normal. In particular, the sample mean taken from a normal population has mean equal to 'm' and standard deviation equal to s/sqrt(n) [sqrt() means square root of], where m is the population mean, s is the population standard deviation and n is the sample size. Although, we don't know the population mean (that's what ...
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