an example showing here there are saving and loan averages about $100000 in depo
ID: 3322585 • Letter: A
Question
an example showing here there are saving and loan averages about $100000 in deposits per week. However, because of the way pay periods fall, seasonality , and erratic fluctuations in the local economy, deposits are subject to a wide variability. In the past , the variance for weekly deposits has been about $199996164. In terms that make more sense to mangers, the standard deviation of weekly deposits has been $14142. Shown here are data from a random sample of 13 weekly deposits for a recent period. Assume weekly deposits are normally distributed. Use these data and a =.10 to test to determine whether the variance for weekly deposits has changed
$93000 $135000 $112000
68000 46000 104000
128000 143000 131000
104000 96000 71000
87000
please solve it step by step with calulation and formula do not use Excel or other program
Explanation / Answer
Given that,
population variance (^2) =199996164
sample size (n) = 13
sample variance (s^2)=832089985.24
null, Ho: ^2 =199996164
alternate, H1 : ^2 !=199996164
level of significance, = 0.1
from standard normal table, two tailed ^2 /2 =18.549
since our test is two-tailed
reject Ho, if ^2 o < - OR if ^2 o > 18.549
we use test statistic chisquare ^2 =(n-1)*s^2/o^2
^2 cal=(13 - 1 ) * 832089985.24 / 199996164 = 12*832089985.24/199996164 = 49.926
| ^2 cal | =49.926
critical value
the value of |^2 | at los 0.1 with d.f (n-1)=12 is 18.549
we got | ^2| =49.926 & | ^2 | =18.549
make decision
hence value of | ^2 cal | > | ^2 | and here we reject Ho
^2 p_value =0
ANSWERS
---------------
null, Ho: ^2 =199996164
alternate, H1 : ^2 !=199996164
test statistic: 49.926
critical value: -18.549 , 18.549
p-value:0
decision: reject Ho
we have enough evidence to support the claim that the variance for weekly deposits has changed
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