Question

You are hired by Google to research how much people are willing to pay for a new cell phone in US. They

are especially interested to know if their new phone, Pixel, should be priced similarly to Apple’s iPhone.

Google believes that there is a difference between what Android and iPhone users are willing to pay for high-end phones. You are hired to answer this question.

To analyze iPhone users your team randomly selects 16 individuals. The sample average for iPhone users is $1100. The sample standard deviation is $400.

a) Find the 95% confidence interval for the average phone price iPhone users are willing to pay. How do you interpret it?

To study Android users you collect data on 25 randomly chosen people. The sample average for Android users is $850. The sample standard deviation is $350.

b) Find the 95% confidence interval for the average phone price that an Android user willing to pay. How do you interpret it.

c) Google executive believes that the price of the phone should be $800. Based on the data forAndroid users can you reject the executive’s hunch?

d) Test the hypothesis that the price iPhone users would pay for an iPhone is the same as the price of an Android user would pay for the phone, against the alternative that they are different. Google uses alpha (?) of 5%. (Hint: This is similar to the situation of comparing house price in Chicago against NYC)

Answer 1

To analyze iPhone users your team randomly selects 16 individuals. The sample average for iPhone users is $1100. The sample standard deviation is $400.

a) Find the 95% confidence interval for the average phone price iPhone users are willing to pay. How do you interpret it?

Sample size = 16.

Sample standard deviation = $400

Sample Mean = $1100

This simple confidence interval calculator uses a t statistic and the sample mean ( $\bar{X}$ ) to generate an interval estimate of a population mean ( $\mu$ ).

The formula for estimation is:

$CI = \bar{X} \pm t_{\alpha/2}\sqrt{\frac{s^2}{n}}$

T score for alpha = 0.05 as significance is 95%. And we divide alpha by 2 to account for 2.5% and 97.5% on either side of the distribution as described in the diagram below

T value for 0.025 with a degree of freedom of 16-1 = 15, is 2.13

$\\CI =1100 \pm 2.1314\sqrt{\frac{400^2}{16}} = 1100 \pm 2.1314*100 = 1100 \pm 213.14\\ \\886.86 <\mu < 1313.14$

Thus, the population mean of the iPhone users lies between $886.86 and $1313.14 with 95% confidence.

To study Android users you collect data on 25 randomly chosen people. The sample average for Android users is $850. The sample standard deviation is $350.

b) Find the 95% confidence interval for the average phone price that an Android user willing to pay. How do you interpret it.

Sample size = 25

Sample standard deviation = $350

Sample Mean = $850

T value for 0.025 with a degree of freedom of 25-1 = 24, is 2.0638

$\\CI =850 \pm 2.0638\sqrt{\frac{350^2}{25}} = 850 \pm 2.0638*70 = 1100 \pm 144.47\\ \\705.53 <\mu < 994.47$

Thus, the population mean of the Android users lies between $705.53 and $994.47 with 95% confidence.

c) Google executive believes that the price of the phone should be $800. Based on the data forAndroid users can you reject the executive’s hunch?

Null Hypothesis and Alternate Hypothesis

$\\H_0: \mu =800\\ \\H_a: \mu \neq800$

Degree of freedom df: 25-1 = 24

Level of significance alpha = 0.05

T statistic

$t = \frac{\bar{X}-\mu}{s/\sqrt{n}} = \frac{850-800}{350/\sqrt{25}} = 50/70 = 0.71428$

The two-tailed p-value for t score of 0.71428 is 0.481993 which is more than significance value of 0.05.

Conclusion: Hence Null hypothesis is accepted and alternate hypothesis is rejected. Executive's claim is accepted.

d) Test the hypothesis that the price iPhone users would pay for an iPhone is the same as the price of an Android user would pay for the phone, against the alternative that they are different. Google uses alpha (?) of 5%.

Null Hypothesis and Alternate Hypothesis

$\\H_0: \mu_i =\mu_a\\ \\H_a: \mu_i \neq \mu_a$

Degree of freedom: Since they have unequal variance and unequal sample size the degree of freedom will be calculated using the following formulae

$df = \frac{[\frac{s_i^2}{n_i}+\frac{s_a^2}{n_a}]^2}{\frac{(s_i^2/n_i)^2}{n_i-1}+\frac{(s_a^2/n_a)^2}{n_a-1}}$

$df = \frac{[\frac{400^2}{16}+\frac{350^2}{25}]^2}{\frac{(400^2/16)^2}{16-1}+\frac{(350^2/25)^2}{25-1}} =28.9 \approx 28$

Level of significance alpha = 0.05

Unpaired t-test statistic

$t = \frac{\bar{x_i}-\bar{x_a}}{\sqrt{s_i^2/n_i+s_a^2/n_a}} = \frac{1100-850}{\sqrt{400^2/16+350^2/25}} = 2.04808$

The two-tailed p-value for t score of 0.71428 is 0.05 which is equal significance value of 0.05 but not less than.

Conclusion: Hence Null hypothesis is accepted and alternate hypothesis is rejected. Concluding, that the price iPhone users would pay for an iPhone is the same as the price of an Android user would pay for the phone.