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Task 1: Solve the next problems:
A student of the writer surveyed her friends and found that amongst 20 males, 4 smoke cigarettes and amongst 30 girl friends, six smoke. Provide two reasons why these benefits should not be intended for a speculation test of the claim that the proportions of male people who smoke and and female people who smoke and are the same. First, the sample is usually not unique, it is a comfort sample and so cannot be reliable. Second, npÒ„ = back button = four < 5, therefore the normal division cannot be utilized to approximate the binomial circulation.
Given a simple random sample of men and an easy random sample of women, we wish to use a 0. 05 significance level to evaluate the claim the fact that percentage of men who also smoke is usually equal to the percentage of women who also smoke. One approach is by using the P-value method of hypothesis testing; the second approach is to use the traditional way of hypothesis tests; and another approach is to base the conclusion on the 95% confidence time period estimate of p1″p2.
Will all approaches often result in the same conclusion?
Make clear. As long as the test is two tailed (so that a confidence interval may be used) and since α =. 05 and percent self confidence = 1 “. 05 =. 96, then yes, the outcomes will USUALLY be the same between traditional approach and confidence time period approach. But it will OFTEN be similar between the p-value approach as well as the confidence period approach. The reason that I declare “almost always is that the regular error solution for pÒ„1-pÒ„2 is slightly different for the confidence period approach than for the hypothesis screening approach. The end result will ALWAYS be similar between the classic and p-value approaches if the test can be two tailed or certainly not.
Criteria pertaining to rejecting Ho:
Traditional way: α =. 05, deny Ho if z < -1.96 or z >1 . ninety six P-value approach: Reject Ho if p-value < .05 (which only occurs if z < -1.96 or z > 1.96). Confidence interval approach: Reject Ho if 0 is not within the confidence interval.
Task 2: Solve the following problems:
The mean tar content of a simple random sample of 25 unfiltered king-sizecigarettes is 21.1 mg, with a standard deviation of 3.2 mg. The mean tar content of a simple random sample of 25 filtered 100 mm cigarettes is 13.2 mg with a standard deviation of 3.7 mg. Assume that the two samples are independent, simple, random samples, selected from normally distributed populations. Do not assume that the population standard deviations are equal. o Use a 0.05 significance level to test the claim that unfiltered king-size cigarettes have mean tar content greater than that of filtered 100 mm cigarettes.
Ho: 1 ¤ 2
Ha: 1 > 2
α = .05
df = nsmaller ” 1 = 25-1 = 24.
Reject Ho if t >1.711
n1 = 25, xÌ…1 = 21.1, s1 = 3.2
n2 = 25, xÌ…2 = 13.2, s2 = 3.7
t = (xÌ…1-xÌ…2)/š[s12/n1+s22/n2] = (21.1-13.2)/š[3.22/25+3.72/25] = 8.075. Since 8.075 >1 ) 711, after that reject Ho.
There is satisfactory evidence to back up the claim that the mean tar content intended for king size smokes is higher than that of strained 100 millimeter cigarettes. um What does the result suggest about the effectiveness of cigarette filters? It appears that cigarette filters are effective in reducing tar. Listed below are systolic blood pressure measurements (mm Hg) obtained from the right and left arms of the same girl. Use a 0. 05 relevance level to check for a big difference between the measurements from the two arms. So what do you determine? Assume that the paired sample data are simple random examples and that the dissimilarities have a distribution that is approximately usual.
Right Provide Left Arm
102 175
101 169
94 182
79 146
79 144
Ho: d = zero
Anordna: d 0
α sama dengan. 05
Reject Ho if p-value <. 05
t-Test: Combined Two Sample for Means Right Arm Left Arm
Mean 91 163. two
Variance 129. a few 297. six
Observations 5 five
Pearson Correlation zero. 867087288 Hypothesized Mean Difference 0 df4 t Stat -17. 33854504 P(T<=t) one-tail a few. 24714E-05 capital t
Important one-tail installment payments on your 131846782 P(T<=t) two-tail 6. 49427E-05 t Crucial two-tail installment payments on your 776445105 to = -17. 338
P-value =. 0000
Seeing that. 000 <. 05, then deny Ho.
There is sufficient evidence to indicate a significant difference in indicate systolic stress between the two arms.
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