PLEASE HELP ASAP I ONLY HAVE 15 MINUTES LEFT
1 answer:
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Answer:
Test statistic is [(n - 1) *S^2 ]/ σ ^2 = [(22 - 1) *(3.9)^2 ]/ (3.4) ^2
with 21 degrees of freedom
Yes this data fits at the 10% level of significance, so I would not reject that statistic of 3.9 mmHg as a wrong standard deviation
Step-by-step explanation:
use the expression I attached in the image to find
[(n - 1) *S^2 ]/ σ ^2
where S = the standard deviation calculated from the sample of n trials.
sigma is the population standard deviation.
[(22 - 1) *(3.9)^2 ]/ (3.4) ^2 = 21 * 15.21 / 11.56 = 27.6306
all we have to do now is to make sure this number is in the 90 % confidence
interval. remember this has 21 degrees of freedom, look at the chi-squared chart.
11.5913 < 27.6306 < 32.67905
where 11.5913 is the lower bound of the chart
and 32.67905 is the upper bound
