Please use TI-84 or excell to solve this problem and show me step by step how yo

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Please use TI-84 or excell to solve this problem and show me step by step how you did it on (the calculator or excel) so I can understand the problem better. thank you

a hypothesis test: State Ho and H,j show what calculation you're using; the conclusion for HO, and finally the conclusion in words. The Freshman 15 is a suspicion that students in college tend to gain 15 pounds or more in their first year. The data below compares September and April weights for 9 randomly selectec incoming students. At the 0.10 significance level, test the claim that the median weight difference is in fact zero, using the Sign Test Sept. wts April wts 122 128 116 114 185 199 154 154 161 165 125 122 122 121 148 149 138 141 The data below is the Detroit Tigers' results for the first half of the month of September 2017. At the 0.05 significance level, use the Sign Test to test the claim that the median difference in runs between teams was zero. Runs Runs Gm# September Opp w/L Detroit OPP 3 2 0 2 133 Friday,Sep 1 (1)CLE 134 Friday Sep 1 (2 CLE 135 Saturday, Sep 2 CLE 136 Sunday, Sep 3 137 Monday, Sep 4 138 Tuesday, Sep5 KCR 139 Wednesday, Sep 6 KCR 140 Friday Sep 8 141 Saturday, Sep 9 IORL 142 Sunday. Sep 10 ITOR 10 6 13 2 KCR 13 TOR 2 8 …… 143 Monday Sep11 CLE 0 144 Tuesday, Sep 12 CLE L 145 Wednesday, Sep 13 CLE 146 Thursday, Sep 14 CHW 17

Explanation / Answer

Here, Null hypothesis is : Their is Weight differences are zero.

And , Alternative Hypotheis is : Their is Weight differences are not zero.

We first Derive difference sign as

Here n= non zero obesrevations are = 8

And x= number positive differences = 3

We determine the probability of obseving X positive differences for a B(n, 1/2) distribution. And use these probability as a p-value for the null hypothesis.

Because the test is one-sided, a result as extreme or more extreme than 3 positive differences includes the results of 3, 4, 5, 6, 7, 8 positive differences,

So we compute the probability as using syntax =BINOMDIST(number_s, trials, probability, pmf)

p(x=3) = 0.2188

p(x=4) = 0.2734

p(x=5) = 0.2188

p(x=6) = 0.1094

p(x=7) = 0.0313

p(x=8) = 0.0039

Sum of above prob. is as = p-value= 0.8555 i.e. p-value > 0.1 (significance level)

This value indicates that there is not strong evidence against the null hypothesis, so we reject null hypothesis

i.e.  Their is Weight differences are not zero.

2)

Same here

Here n= non zero obesrevations are = 14

And x= number positive differences = 2

We determine the probability of obseving X positive differences for a B(n, 1/2) distribution. And use these probability as a p-value for the null hypothesis.

Because the test is one-sided, a result as extreme or more extreme than 2 positive differences includes the results of 2,3, 4, 5, 6, 7, 8,9,10,11,12,13,14 positive differences,

So we compute the probability as using syntax =BINOMDIST(number_s, trials, probability, pmf)

p(x>=2) = 1 - p(x<2) = 1 - (p(x=o) + p(x = 1))

= 1 - 0.000061 + 0.00085

= 1 - 0.00092

= 0.9991

Sum of above prob. is as = p-value= 0.9991 i.e. p-value > 0.1 (significance level)

This value indicates that there is not strong evidence against the null hypothesis, so we reject null hypothesis

i.e.  Their is run differences between teams are not zero.

obs Sept. Wt. April Wt. Diff. Sign 1 122 128 - 2 116 114 + 3 185 199 - 4 154 154 0 5 161 165 - 6 125 122 + 7 122 121 + 8 148 149 - 9 138 141 -

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