Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day
1 answer:
Answer:
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
Step-by-step explanation:
Given that a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2.
Group Group One Group Two
Mean 75.8600 75.4100
SD 16.9100 19.7300
SEM 2.8583 3.2436
N 35 37
*SEM is std error/sqrt n
Mean difference = 0.4500
(two tailed test)
Std error for difference = 4.342
Test statistic t =
df =70
p value = 0.9178
Since p >0.05 we accept H0
c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2
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The question is incomplete. The complete question is :
In a survey of women in a certain country ( ages 20-29), the mean height was 62.9 inches with a standard deviation of 2.81 inches. Answer the following questions about the specified normal distribution. (a) What height represents the 99th percentile? (b) What height represents the first quartile? (Round to two decimal places as needed)
Solution :
Let the random variable X represents the height of women in a country.
Given :
X is normal with mean, μ = inches and the standard deviation, σ =
inches
Let,
, then Z is a standard normal
a). Let the percentile is = a
The point a is such that,
From standard table, we get :
∴
= 6.536903 + 62.9
= 69.436903
= 69.5 (rounding off)
Therefore, the height represents the percentile = 69.5 inches.
b). Let b = height represents the first quartile.
It is given by :
From the standard normal table,
∴
= 1.895345 + 62.9
= 64.795345
= 64.8 (rounding off)
Therefore, the height represents the 1st quartile is 64.8 inches.