Questions: Hi-def: Following are prices of a random sample of 17 smart TVs sold on shopper.cnet.com in 2013 with screen sizes between 46 and 50 inches, along with dotplot of the data. 598, 697, 699, 749, 799, 829, 849, 928, 1050, 1098, 1169, 1198, 1269, 1299, 1399, 1455, 1599 (a) Is it reasonable to assume that the conditions for constructing a confidence interval for the mean price are satisfied? It is reasonable to assume the conditions are satisfied. (b) If appropriate, construct a 95% confidence interval for the mean price of all smart TVs in this size range. Round the answers to two decimal places. A 95% confidence interval for the mean price of all smart TVs in this size range is 884.00 < μ < 1184.24.

Solution Steps

To construct a confidence interval for the mean price, we need to check if the conditions are satisfied:

1. Random Sample: The data is a random sample of 17 smart TVs.
2. Normality: For small sample sizes (n < 30), the data should be approximately normally distributed. Given the dot plot, the data appears to be roughly symmetric without extreme outliers.

Calculate the sample mean (\(\bar{x}\)) and sample standard deviation (s) of the given prices.

Prices: 598, 697, 699, 749, 799, 829, 849, 928, 1050, 1098, 1169, 1198, 1269, 1299, 1399, 1455, 1599

\[ \bar{x} = \frac{\sum x_i}{n} = \frac{598 + 697 + 699 + 749 + 799 + 829 + 849 + 928 + 1050 + 1098 + 1169 + 1198 + 1269 + 1299 + 1399 + 1455 + 1599}{17} = 1044.12 \]

\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{(598-1044.12)^2 + (697-1044.12)^2 + \ldots + (1599-1044.12)^2}{16}} = 292.92 \]

Use the t-distribution for the confidence interval since the sample size is small (n < 30).

Degrees of freedom (df) = n - 1 = 17 - 1 = 16

The critical value (t*) for a 95% confidence interval with 16 degrees of freedom is approximately 2.120.

\[ \text{Margin of Error} = t^* \times \frac{s}{\sqrt{n}} = 2.120 \times \frac{292.92}{\sqrt{17}} = 150.12 \]

\[ \text{Confidence Interval} = \bar{x} \pm \text{Margin of Error} = 1044.12 \pm 150.12 \]

Final Answer