It would be <span>b.<span>$1,937.12.
Thank you! I am sorry if this was a bit late.
</span></span>
<span>Table:
Class Boundaries Frequency
5-10 8
10-15 9
15-20 15
20-25 10
25-30 8
30-35 6
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total 56
Average =
[5+10]/2*8+[10+15]/2*9+[15+20]/2*15+[20+25]/2*10+[25+30]/2*8+[30+35]/2*6
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</span> 56
That is 1075 / 56 = 19.2
Answer: 19
Answer: We should reject the null if the test statistic is greater than <u>1.895</u>.
Step-by-step explanation:
We assume the population to be normally distributed.
Given: Sample size : 
, which is asmall sample (n<30), so we use t-test.
We always reject the null hypothesis if the absolute t-value is greater than critical value.
Therefore, We should reject the null if the test statistic is greater than <u>1.895</u>.
Answer:
(Frequency of the value being measured divided by the total frequency) times by 100
The correct answers are A, D, and E.
