Answer:
Step-by-step explanation:
a) For west coast customers, Confidence interval is written in the form,
(Sample mean - margin of error, sample mean + margin of error)
The sample mean, x is the point estimate for the population mean.
Margin of error = z × s/√n
Where
s = sample standard deviation = 0.52
n = number of samples = 174
From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score
In order to use the t distribution, we would determine the degree of freedom, df for the sample.
df = n - 1 = 174 - 1 = 173
Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05
α/2 = 0.05/2 = 0.025
the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975
Looking at the t distribution table,
z = 1.9738
Margin of error = 1.9738 × 0.52/√174
= 0.078
the lower limit of this confidence interval is
3.24 - 0.078 = 3.162
the upper limit of this confidence interval is
3.24 + 0.078 = 3.318
Therefore, we are 95% confident that the population mean score for west coast customers is between 3.162 and 3.318
b) Confidence interval for the difference in the two population means is written as
Difference in sample mean ± margin of error
Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)
Where
x1 = sample mean of east coast customers
x2 = sample mean of west coast customers
s1 = sample standard deviation of east coast customers
s2 = sample standard deviation of west coast customers
n1 = number of east coast customers surveyed
n2 = number of west coast customers surveyed
For a 95% confidence interval, the z score is 1.96
From the information given,
x1 = 3.51
s1 = 0.51
n1 = 174
x2 = 3.24
s2 = 0.52
n2 = 355
x1 - x2 = 3.51 - 3.24 = 0.27
Margin of error = 1.96√(0.51²/174 + 0.52²/355) = 1.96√0.00225651773
= 0.093
Confidence interval = 0.27 ± 0.093
c) This is a test of 2 independent groups. Let μ1 be the mean score of east coast customers and μ2 be the mean score of wet coast customers.
The random variable is μ1 - μ2 = difference in the mean score between east coast and west coast customers
We would set up the hypothesis.
The null hypothesis is
H0 : μ1 ≥ μ2 H0 : μ1 - μ2 ≥ 0
The alternative hypothesis is
H1 : μ1 < μ2 H1 : μ1 - μ2 < 0
Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is
(x1 - x2)/√(s1²/n1 + s2²/n2)
t = (3.51 - 3.24)/√(0.51²/174 + 0.52²/355)
t = 5.68
The formula for determining the degree of freedom is
df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²
df = [0.51²/174 + 0.52²/355]²/[(1/174 - 1)(0.51²/174)² + (1/355 - 1)(0.52²/355)²] = 0.00000509187/0.00000001456
df = 350
We would determine the probability value from the t test calculator. It becomes
p value < 0.00001
This is very close to 0
Since the p value is very small, there is a very high possibility of rejecting the null hypothesis. It means that the difference in the mean score between east coast and west coast customers is less than zero. Therefore, the average score of the east coast customers is not higher than the average score of the west coast
