Mathematics, 02.11.2020 16:30, beverly3966

The mean cost of a meal for two in a mid-range restaurant in Tokyo is $40 (Numbeo 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.

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answered: lliaaaa

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Question

The mean cost of a meal for two in a mid-range restaurant in Tokyo is $40 (Numbeo website, December 14, 2014). How do prices for comparable meals in Hong Kong compare? The DATAfile Hong Kong Meals contains the costs for a sample of 42 recent meals for two in Hong Kong mid-range restaurants

a. 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?

The margin of error is E = 2.13

The 95% confidence interval is $30.53 < \mu < 34.79$

The cost of mean in Hong Kong is lower

Step-by-step explanation:

From the question we are told that

The mean cost of a meal for two in a mid-range restaurant in Tokyo is $\mu_1 = \$ 40$

Generally the mean for a meal in Hong Kong is mathematically represented as

$\= x = \frac{\sum x_i }{n}$

=> $\= x = \frac{ 22.78+ 33.89 +\cdots + 37.93 }{42}$

=> $\= x =32.66$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{\frac{ \sum (x_i - \= x )^2}{n-1} }$

=> $\sigma = \sqrt{\frac{ (22.78 - 32.66 )^2+ (33.89 - 32.66 )^2+\cdots + (37.93 - 32.66 )^2}{42-1} }$

=> $\sigma = 6.83$

Generally the degree of freedom is mathematically represented as

df = n- 1

=> df = 42 - 1

=> df = 41

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the t distribution table the critical value of $\frac{\alpha }{2}$ at a degree of freedom of df = 41 is

$t_{\frac{\alpha }{2} , 41} = 2.020$

Generally the margin of error is mathematically represented as

$E = t_{\frac{\alpha }{2} ,41 } * \frac{\sigma }{\sqrt{n} }$

=> $E = 2.0203 * \frac{6.83 }{\sqrt{42} }$

=> E = 2.13

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $32.66 - 2.13 < \mu < 32.66 + 2.13$

=> $30.53 < \mu < 34.79$

Generally comparing the mean cost of a meal for two in a mid-range restaurant in Tokyo to the 95% confidence interval estimate of the population mean cost of mean in Hong Kong we see that the cost of meals in Hong Kong is lower.