Question

Assume that the weights of adult women are normally distributed, and that the population standard deviation of weights of all women is o = 29 lb. A researcher claims that the weights of supermodels vary less than the weights of women in general. To test the claim, a random sample of nine supermodels is selected, with data (in pounds) shown as below: 125 119 128 128 119 127 105 123 115 Using a .05 significance level, does this data support the researcher's claim? The test statistic is.54, which is less than the critical value. Therefore, there is statistical evidence that supermodels' weights are less variable than the general population. The sample SD is less than 29 lbs, which means that supermodels' weights vary less than the general population. The test statistic is.54, which is less than the critical value. Therefore, we cannot say that supermodels' weights are less variable than the general population. The test statistic is.54, which is greater than the significance level. Therefore, we cannot say that supermodels' weights are less variable than the general population. 1) The table below shows the number of students from three different high schools participating in the mathematics and physics sections of a science fair: Number of Students High School 1 High School 2 High School 3 8 9 12 28 20 19 Mathematics Physics Using a .05 level of significance, test the claim that the section of participation and the high school where the students were from are independent. There is evidence to reject the claim that the high school and the section of participation are independent because the test value 5.991> 0.821 There is not evidence to reject the claim that the high school and the section of participation are independent because the test value 2.161 < 5.991 There is not evidence to reject the claim that the high school and the section of participation are independent because the test value 0.821 < 5.991 There is evidence to reject the claim that the high school and the section of participation are independent because the test value 12.592 > 0.821 b)

please help with both, thank you very much!

Answer 1

Solution-1:

Ho:sigma=29

Ha:sigma<29

For the given sample ,n=9

sample standard deviation=7.53326

F=(n-1)*s^2/sigma^2

=(9-1)* 7.53326^2/29^2

=0.5398336

chi sq critcal value

=CHISQ.INV(0.05,8)

=2.732636793

answer:

test statistic is less than the critical value.Therefore,we cannot say that supermodel's weights are less variable than the general popualtion

OPTION C

Solution-2:

Ho:indpendent

Ha:dependent

TEXAS INSTRUMENTS TI-83 Plus X2-Test X2=2.161115713 P=.3394061329 df=2 MATRIX[A] 2 X3 [B 9 12 [ 28 20 19 2, 3=19 FORMAT 3 CALC F4 TABLE STATPLOTF: TESET P2 YE WINDOW ZOOM TRACE GRAPH STAT PLOT FETISET 2 FORMAT F9 TABLE F5 CALC Ft TRACE Y= WINDOW ZOOM GRAPH INS QUIT MODE 2nd DEL A-LOCK ALPHA LINK Xten LIST STAT

chi sq=2.161

chi sq critical vale

=CHISQ.INV.RT(0.05,2)

=5.991464547

p=0.3394

ANSWER:

There is not evidence to reject the claim that the high school and section of participants are independent because

the test value 2.161<5.991

OPTION B