Answer:
Step-by-step explanation:
sample size = n = 60
Sample mean = x bar = 5.25
Sample sd = s = 0.55
Std error of sample = s/sqrt n = 0.071
(One tailed test)
Mean difference = 5.25-5.4 = -0.15
Test statistic t = Mean diff/Se = -2.11
df=59
(since population std dev is not known t test is used)
p value =0.01955
Since p <alpha =0.05 we reject null hypothesis.
The results suggest about the sample group that the sample is from a population with a mean less than 5.4
Answer:
a) Sample mean = 9.99
Sample standard deviation = 0.3348
b) -1.0389
c) 0.1631
Step-by-step explanation:
We are given the following in the question:
9.7, 9.9, 10.3, 10.1, 10.5, 9.4, 9.9, 10.1, 9.7, 10.3
a) sample mean and standard deviation
Formula:
where are data points,
is the mean and n is the number of observations.
Sum of squares of differences = 1.009
b) observed value of the t-statistic
Formula:
Putting all the values, we have
c) probability of these statistics (or worse) if the true mean were 10.1 mm
Degree of freedom = n - 1 = 9
Calculating the value from the table
0.1631 is the the probability of these statistics (or worse) if the true mean were 10.1 mm
