Question

ch 20 nine randomly sampled students were asked how many hours of tv they watched last week. \\( \overline { x } = 3.5 \mathrm { h } \\) and \\( s = 4 \mathrm { h } \\). for \\( 95 \\% \\) confidence and 8 degrees of freedom, \\( t ^ { * } = 2.3 \\). the stemplot was roughly symmetrical, with no outliers. what is the \\( 95 \\% \\) confidence interval for \\( m \\)?
\\( 3.5 \pm ( 2.3 \times 4 / 3 ) \\)
\\( 3.5 \pm ( 2.3 \times 4 \times 9 ) \\)
\\( 3.5 \pm ( 2.3 \times 4 / 9 ) \\)
\\( 3.5 \pm ( 2.3 \times 4 ) \\)
question 2
ch 20 the time (in number of days) until maturity of a certain variety of tomato plant is normally distributed. i select a simple random sample of four plants of this variety and measure the time until maturity. the sample yields xbar \\( = 65 \\) with standard deviation \\( s = 24 \\). you read on the package of seeds that these tomatoes reach maturity, on average, in 61 days. you want to test to see if your seeds are reaching maturity later than expected, which might indicate that your package of seeds is too old. the appropriate hypotheses are
\\( h o : m u = 65, h a : m u < 65 \\).
\\( h o : m u = 65, h a : m u > 65 \\).
\\( h o : m u = 61, h a : m u > 61 \\).
\\( h o : m u = 61, h a : m u < 61 \\).

Explanation:

Step1: Recall the formula for confidence interval

The formula for a confidence interval for the population mean \( \mu \) when the population standard deviation \( \sigma \) is unknown is \( \bar{x}\pm t^{*}\frac{s}{\sqrt{n}} \). Here, \( \bar{x} = 3.5\), \(t^{*}=2.3\), \(s = 4\), and \(n = 9\) (since there are 9 students sampled). The standard error is \( \frac{s}{\sqrt{n}}=\frac{4}{\sqrt{9}}=\frac{4}{3}\).

Step2: Substitute values into the formula

Substituting the values into the formula \( \bar{x}\pm t^{*}\frac{s}{\sqrt{n}}\), we get \(3.5\pm(2.3\times\frac{4}{3})\).

Answer:

\(3.5\pm(2.3\times\frac{4}{3})\) (the first option)