Question

Your parents want to understand whether bottle rockets using pure vida water bottles shoot higher than the average home science experiment water bottles. They find on the home-schooled list serve that the average bottle rocket flies about 150 feet high. There is no standard deviation given on this list serve. These parents of yours test 8 bottle rockets with the pure vida water bottles. They get 151, 157, 148, 164, 152, 149, 153, 153.

1. Is this a one-tailed or two-tailed test?
2. State the null and alternative hypotheses.
3. Calculate tobt . Recall that you will need to first calculate S ¯ = √SN .
4. What is tcv for a 95 percent level of confidence?
5. Should the null hypothesis be rejected?
6. Construct a 95 percent confidence interval for X¯ .

Answer 1

1. This is a one-tailed test.

2. Null hypothesis (
`\(H_0\))`: The average height of bottle rockets using pure vida water bottles is equal to or less than 150 feet. Alternative hypothesis
`(\(H_a\))`: The average height of bottle rockets using pure vida water bottles is greater than 150 feet.

`3. \(t_{\text{obt}} = 1.64\)`

4.
`\(t_{\text{cv}} = 1.833\)`

5. The null hypothesis should not be rejected.

6. The 95 percent confidence interval for the average height of bottle rockets using pure vida water bottles is
`\(145.9 \text{ feet} < \mu < 157.5 \text{ feet}\).`

Step-by-step explanation:

In a one-tailed test, we are interested in whether the observed data falls in only one direction from the mean. In this case, the parents are investigating whether the pure vida water bottle rockets fly higher than the average, making it a one-tailed test.

The null hypothesis
`(\(H_0\))` posits that the average height of bottle rockets using pure vida water bottles is equal to or less than 150 feet. The alternative hypothesis
`(\(H_a\))` suggests that the average height is greater than 150 feet.

To calculate
`\(t_{\text{obt}}\),` we first find the standard deviation (S of the sample and then divide it by the square root of the sample size n. The obtained
`\(t_{\text{obt}}\)`is then compared with the critical value
`(\(t_{\text{cv}}\))` for a 95 percent confidence level. If
`\(t_{\text{obt}}\)`is less than
`\(t_{\text{cv}}\),` we fail to reject the null hypothesis.

The critical value
`(\(t_{\text{cv}}\))` for a one-tailed test with 7 degrees of freedom at a 95 percent confidence level is 1.833. Since
`\(t_{\text{obt}}\)` is less than
`\(t_{\text{cv}}\),`we do not have enough evidence to reject the null hypothesis.

Finally, the 95 percent confidence interval for the average height is calculated using the formula
`\(\bar{X} \pm t_{\text{cv}} * (S)/(√(n))\).` This interval provides a range within which we are 95 percent confident that the true average height lies.