Respuesta :
Answer:
a) [tex]Me=2.02 \frac{6.827}{\sqrt{42}}=2.13[/tex]
b) So on this case the 95% confidence interval would be given by (30.53;34.79)
c) On this case the confidence interval not contains the price $40, so we can conclude that the prices for Hong Kong mid-range restaurants are significant less than the prices for mid range restaurants in Tokyo at 5 % of significance.
Step-by-step explanation:
1) Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n represent the sample size
2) Part a
The margin of error is given by:
[tex]Me=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=42-1=41[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,41)".And we see that [tex]t_{\alpha/2}=2.02[/tex]
[tex]Me=2.02 \frac{6.827}{\sqrt{42}}=2.13[/tex]
3) Part b
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
The mean calculated for this case is [tex]\bar X=32.66[/tex]
The sample deviation calculated [tex]s=6.827[/tex]
Now we have everything in order to replace into formula (1):
[tex]32.66-2.02\frac{6.827}{\sqrt{42}}=30.532[/tex]
[tex]32.66+2.02\frac{6.827}{\sqrt{42}}=34.788[/tex]
So on this case the 95% confidence interval would be given by (30.532;34.788)
4) Part d
On this case the confidence interval not contains the price $40, so we can conclude that the prices for Hong Kong mid-range restaurants are significant less than the prices in mid range restaurants in Tokyo at 5 % of significance.
a) The margin of error is =2.13.
b) The 95% confidence interval estimate of the population mean is 34.788 and 30.532.
c) The prices for Hong Kong mid-range restaurants are significant less than the prices in mid range restaurants in Tokyo at 5 % of significance.
What is meant by the margin of error?
A margin of error is a statistical measurement that accounts for the difference between actual and projected results in a random survey sample.
a) The margin of error is given by:
[tex]M_e=t_{\alpha /2} \frac{s}{\sqrt{n} }[/tex]
Now, degree of freedom
f= n-1
= 42-1
=41
As, the level of confidence is 95 % , α=0.05 and α/2= 0.025
[tex]t_{\alpha /2}[/tex]=2.02
So, Me= 2.02* 6.827/√42
=2.13.
b) The confidence interval for the mean is ,
[tex]X\pm t_{\alpha /2}\frac{s}{\sqrt{n} }[/tex]
so, the mean will be 32.66 and deviation will be 6.827
Hence, 32.66+2.02* 6.827/√42=34.788
and, 32.66-2.02* 6.827/√42=30.532
c)In this case the confidence interval not contains the price $40.
so, conclude that the prices for Hong Kong mid-range restaurants are significant less than the prices in mid range restaurants in Tokyo at 5 % of significance.
Learn more about margin of error here:
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