In a certain school district, it was observed that 29% of the students in the el
ID: 3368121 • Letter: I
Question
In a certain school district, it was observed that 29% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 91 out of 255 students are only children. The school district administrators want to know if the proportion of only children in the special program is significantly different from the proportion for the school district. Test at the ? = 0.02 level of significance.
Using the normal approximation for the binomial distribution (without the continuity correction), was is the test statistic for this sample based on the sample proportion? Report answer as a decimal accurate to 3 decimal places.) You are now ready to calculate the P-value for this sample. P-value Report answer as a decimal accurate to 4 decimal places.) This P-value (and test statistic) leads to a decision to... reject the null accept the null fail to reject the null reject the alternative As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T; program. There is not sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T; program. The sample data support the assertion that there is a different proportion of only children in the G&T; program. There is not sufficient sample evidence to support the assertion that there is a different proportion of only children in the G&T; program.Explanation / Answer
p(hat)= 91/255=0.35686
n= 255
H0: P= 0.29
Ha: P? 0.29 .....two tail test
alpha= 0.02
Test Statistics
Z= (p(hat)-P)/sqrt(p*q/n)
=(0.35686-0.29)/sqrt(0.29*0.71/255)
=2.3530
P-value=1-P(Z<2.3530)= 2*0.00931 (this p-value is for one tail to get p-value for two tail we need to multiply it by 2.)
rounded p-value= 0.01862
Decision making
P-value ? Alpha Reject H0
0.01862 <0.02
there is sufficient evidence to warrent rejection of the assertion that there is a different proportion of only children in the G & T program.
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answers.
Z=2.3530
p-value=0.01862
Reject null hypothesis
there is sufficient evidence to warrent rejection of the assertion that there is a different proportion of only children in the G & T program.
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