Mathematics, 06.12.2019 19:31 miya214

The mean cost of a meal for two in a mid-range restaurant in tokyo is $40 (numbeo. com website, december 14, 2014). how do prices for comparable meals in hong kong compare? the datafile hongkongmeals contains the costs for a sample of 42 recent meals for two in hong kong mid-range restaurants. data: meal price ($)22.7833.8922.7718.0423.2935.2842 .3836.8838.5541.6825.7334.1931.7525 .2426.3219.5736.5732.9736.8330.1737 .2925.3724.7128.7932.8343.0035.2334 .7633.0627.7331.8938.4739.4240.7243 .9236.5145.2533.5129.1730.5426.7437 .93a. with 95% confidence, what is the margin of error (to 2 decimals)?
b. what is the 95% confidence interval estimate of the population mean (to 2 decimals)?
c. how do prices for meals for two in mid-range restaurants in hong kong compare to prices for comparable meals in tokyo restaurants?

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Answer from: am2garcia5

a) $Me=2.02 \frac{6.827}{\sqrt{42}}=2.13$

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".

$\bar X$ represent the sample mean for the sample

$\mu$ 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:

$Me=t_{\alpha/2}\frac{s}{\sqrt{n}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

df=n-1=42-1=41

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , 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 $t_{\alpha/2}=2.02$

$Me=2.02 \frac{6.827}{\sqrt{42}}=2.13$

3) Part b

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=32.66$

The sample deviation calculated s=6.827

Now we have everything in order to replace into formula (1):

$32.66-2.02\frac{6.827}{\sqrt{42}}=30.532$

$32.66+2.02\frac{6.827}{\sqrt{42}}=34.788$

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.

Answer from: Quest

correct answer:

b.(19,-2)

Answer from: Quest

i think the amount would be $140.62

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