Question 1012619: Please please please can someone help me thank you so much please!!!!!!!
In a study of pregnant women and their ability to correctly predict the sex of their baby, 59 of the pregnant women had 12 years of education or less, and 40.7% of them correctly predicted the sex of their baby. Use a=0.05 significance level to test the claim that these women have no ability to predict the sex of their baby, and the results are not significantly different from those that would be expected with random guesses. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
A. H0: p=0.407 H1: p≠0.407
B. H0: p=0.407 H1: p>0.407
C. H0: p=0.5 H1: p≠0.5
D. H0: p=0.407 H1: p<0.407
E. H0: p=0.5 H1: p<0.5
F. H0: p=0.5 H1: p>0.5
The test statistic is z=______.(Round to two decimal places as needed.)
The P-value is_______.(Round to four decimal places as needed.)
Identify the conclusion about the null hypothesis and the final conclusion that address the original claim.
___________ (Fail to reject, Reject H0). There __________ (is, is not) sufficient evidence to warrant rejection of the claim that these women have no ability to predict the sex of their baby. The results for these women with 12 years of education or less suggests that their percentage of correct predictions ____________ (is not, is very)different from results expected with random guesses.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! with random guesses, you would expect that 50% of the women would be able to predict the sex of thei baby and 50% wouldn't.
that's because the probability of random guesses is fifty fifty.
it's either a boy or a girl.
there are cases where the infant cannot be considered purely boy or purely girl, but these cases are very rare and are not considered in this analysis.
so, it's fifty fifty.
the baby is either a boy or a girl.
if you guessed randomly, then you will probably guess right 50% of the time and guess wrong 50% of the time.
that's the null alternative.
In a study of pregnant women and their ability to correctly predict the sex of their baby, 59 of the pregnant women had 12 years of education or less, and 40.7% of them correctly predicted the sex of their baby. Use a=0.05 significance level to test the claim that these women have no ability to predict the sex of their baby, and the results are not significantly different from those that would be expected with random guesses. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
the problem states that 59 of the pregnant women had 12 years of education or less, and 40.7% of them correctly predicted the sex of their baby.
the alternative hypothesis is that they did not have the capability to predict the sex of their baby.
based on this sample, you are to determine whether the claim that they do not have the capability to predict the sex of their baby has merit, or whether the results of the sample are not sufficiently strong enough for you to say that the claim has merit.
if the sample is within the confidence limits of the data, then you would say that the sample could have resulted from random sample variation and so there isn't strong enough evidence to suggest that these women were actually worse predictors than random chance would allow.
if the sample is outside the confidence limits of the data, they you would say that the probability that the sample could have resulted from random sample variation is so remote that the claim is likely true.
since you are dealing with a proportion, and since a normal distribution of the population is assumed, you would do the following.
p = proportion assumed able to predict = 40.7% / 100 = .407.
q = proportion assumed mot able to predict = 1 - .407 = .593.
n = sample size = 59
the population standard deviation is assumed to be normal.
p is assumed to be equal to .5 for the population.
q is therefore equal to 1 - .5 = .5 for the population.
the sample standard error is calculated as follows:
s = sqrt(p*q/n)
if you know what the population proportion is, you use that.
otherwise you use the proportion from the sample.
in this case, the population proportion is assumed to be .5, so we'll use that.
if in doubt, the logical thing to do is to do the study twice, once with the assumed standard error and once with the sample standard error.
if the results indicate the same result, you're home free.
otherwise, you have some issues to resolve.
i'll do both for you to show you what can happen in this case.
using p = .5 and q = .5, the stnadard error would be calculated as follows:
s = sqrt(p*q/n) becomes s = sqrt(.5*.5/59) = .06509...
using p = .407 and q = .593, the standard error would be calculated as follows:
s = sqrt(p*q/n) becomes s = sqrt(.407*.593/59) = .06395...
you should go with whatever source of reference you are using since you'll be graded based on that.
you would now calculate the z-score of your data.
using the sample proportion in the calculation of s, you would do the following:
z = (x-m)/s
x is the raw score which is equal to .407
m is the mean of the population which is assumed to be .5
s is the standard error of the mean calculated from the sample proportion which is equal to .06395...
you get z = (.407-.5)/.06395... = -1.45406...
using the assumed population proportion in the calculation of s, you would do the following:
z = (x-m)/s
x is the raw score which is equal to .407
m is the mean of the population which is assumed to be .5
s is the standard error of the mean calculated from the sample proportion which is equal to .06509...
you get z = (.407-.5)/ .06509...-1.42869...
at a one tail 95% confidence level (each tail outside the confidence limits is equal to .05), your critical z-factor will be plus or minus 1.64485...
since -1.42869... is less than that, and since -1.45406 is also less than that, it didn't matter for this study whether you used the assumed population proportion or the sample proportion in the calculation of the standard error.
this makes your case stronger and doesn't introduce any questions into the validity of the result.
since the z-factor of the study is less than the critical z-factor, your conclusion is that there is not enough evidence that the 59 women had any special lack of talent for predicting the sex of their baby that was any worse than if they had randomly guessed.
the sample, being within the limits of the confidence level, is assumed to be within the sampling error criteria and therefore can be considered in line with what a random guess would have achieved.
in other words, if they had just guessed randomly, a sample of 59 women guessing randomly, was possible and the results of the sample could have been due to sampling error.
with a population mean of .5, a z-score of -1.64485... would yield a raw score of around .3929... or .3947...
.407 was above that so it was within the limits imposed on the study.
exceed the limits and you can reject the null hypothesis and say the alternate hypothesis is probably more accurate in the study group.
don't exceed the limits and you don't have enough information to reject the null hypothesis and so you assume the alternate hypothesis is not sufficiently different to be due to anything other than random sample variation.
that's the gist of it as far as i understand.
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