Question

a) A magazine collects data each year on the price of a hamburger in a certain fast food restaurant in various countries around the world. The price of this hamburger for a sample of restaurants in Europe in January resulted in the following hamburger prices (after conversion to U.S. dollars).

5.16	4.95	4.09	4.66	5.28	4.69
4.12	4.96	5.15	5.55	5.34	4.60

The mean price of this hamburger in the U.S. in January was $4.61. For purposes of this exercise, assume it is reasonable to regard the sample as representative of these European restaurants. Does the sample provide convincing evidence that the mean January price of this hamburger in Europe is greater than the reported U.S. price? Test the relevant hypotheses using  α = 0.05.

(Use a statistical computer package to calculate the P-value. Round your test statistic to two decimal places and your P-value to three decimal places.)

t= __

p- value = ___

b) The risk of developing iron deficiency is especially high during pregnancy. Detecting such a deficiency is complicated by the fact that some methods for determining iron status can be affected by the state of pregnancy itself. Consider the following data on transferrin receptor concentration for a sample of women with laboratory evidence of overt iron-deficiency anemia.

15.3	9.2	7.7	11.9	10.4	9.7
20.3	9.4	11.4	8.3	9.4	16.1

(a) Compute the values of the sample mean and median. (Round your answers to two decimal places.)

mean	=
median	=

Which of the mean and median do you regard as more representative of the sample, and why?

choice a) The mean is always more representative of a sample.

choice b)The mean because it is not influenced by the three extreme values at the upper end of the distribution.  

choice c) The median because it is not influenced by the three extreme values at the upper end of the distribution.

choice d) The median is always more representative of a sample.

Answer 1

Answer a:

Let be the mean price of hamburger in the U.S. in January.

For testing of the situation, the Null Hypothesis, H0 : = $4.61 against the Alternative Hypothesis, Ha: > $4.61

This is a one sided test (right - tailed)

The following table shows all the calculations -

	Observations(x)	x -	(x - )^2
	5.16	0.28	0.0784
	4.95	0.07	0.0049
	4.09	-0.79	0.6241
	4.66	-0.22	0.0484
	5.28	0.4	0.16
	4.69	-0.19	0.0361
	4.12	-0.76	0.5776
	4.96	0.08	0.0064
	5.15	0.27	0.0729
	5.55	0.67	0.4489
	5.34	0.46	0.2116
	4.60	-0.28	0.0784
Total	58.54		2.3477

The number of observations, n = 12

Mean, = 58.54 / 12 = 4.88

Standard Deviation, s’ = {(x - )^2 / (n - 1)}^0.5 = 0.46

Now, the test statistic is t = [(n^0.5) ( - )]/ s’ which follows t – distribution with (n - 1) degrees of freedom under the null hypothesis.

Substituting all the values,

t = 2.03

Also, p – value = 0.033

The Significance level = 0.05

The critical value, t0.05, 11 = 1.796 (value taken from t – distribution table)

If the value of the test statistic is greater than the critical value, then we reject the null hypothesis, or else we fail to reject the null hypothesis.

Also, if the p – value is less than the significance level, we reject the null hypothesis else one fails to reject the null.

Since, t > The Critical Value, also, p – value < significance level, we reject the null hypothesis

Therefore, the sample does provide convincing evidence that the mean January price of the Hamburger is not equal to $4.61, so it must be greater than that value

• Assumption : We assume tha the sample has been taken from a Normal Distribution

Answer b:
Mean of the Observations, = (15.3 + 9.2 + 7.7 + 11.9 + 10.4 + 9.7 + 20.3 + 9.4 + 11.4 + 8.3 + 9.4 + 16.1) / 12

= 11.59

Arranging all the observations in Ascending Order –

7.7, 8.3, 9.2, 9.4, 9.4, 9.7, 10.4, 11.4, 11.9, 15.3, 16.1, 20.3

Median = (6th Observation + 7th Observation)/2 = (9.7 + 10.4) / 2 = 10.05

The Mean is the average of the given sample, and it represents the centre point of the data values and always has been more representative of a sample.

So, choice a is the most appropriate choice.

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