Mathematics, 18.12.2019 21:31 smhrosepetals

Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. the data follows: m = 5, x = 114.1, s1 = 5.08, n = 5, y = 129.9, and s2 = 5.37. calculate a 95% ci for the difference between true average stopping distances for cars equipped with system 1 and cars equipped with system 2. (round your answers to two decimal places.)

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Answer from: hemaam8

The 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =114.1$ represent the sample mean 1

$\bar X_2 =129.9$ represent the sample mean 2

n1=5 represent the sample 1 size

n2=2 represent the sample 2 size

s_1 =5.08 sample standard deviation for sample 1

s_2 =5.37 sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =114.1-129.9=-15.8$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

df=n_1 +n_2 -1=5+5-2=8

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=\pm 2.31$

The standard error is given by the following formula:

$SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=3.306$

Confidence interval

Now we have everything in order to replace into formula (1):

$-15.8-2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-23.437$

$-15.8+2.31\sqrt{\frac{5.08^2}{5}+\frac{5.37^2}{5}}=-8.163$

So on this case the 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

Answer from: Quest

121 pints. they had 176 and drank 55

Answer from: Quest

which would be true for stem changing verbs?

they have a consistent spelling change in the stem of the verb.

what is different about the verb “venir” as compared to other stem-changing verbs?

only the third person forms have a stem change.

which of the following could not be considered a “stem-changing” verb?

querer

which of the following would not be condsidered an event?

desfile

what word could be used in response to the following?

¿qué quieres comprar?

quero

what word could be used in response to the following?

¿dónde uds. pueden ir?

vamos

ellos ir al centro comercial.

yo comprar un vestido.

pedimos

dónde tú?

dormes

step-by-step explanation:

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Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped wit...