Mathematics, 05.12.2019 23:31 savannahsharp4463

When the background music tempo was slow, the mean amount of bar purchases for a sample of 19 restaurant patrons was $30.44 with a standard deviation of $15.90. when the background music tempo was fast, the mean amount of bar purchases for a sample of 14 patrons in the same restaurant was $21.17 with a standard deviation of $9.10. assuming equal variances, at α = 0.10, is the true mean higher when the music is slow?

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Answer from: Tracic1972

Comparing the p value with a significance level for example $\alpha=0.1$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can't say that the population mean for music tempo low is significantly higher than the population mean when the music is fast.

Step-by-step explanation:

Data given and notation

$\bar X_{A}=30.44$ represent the mean for music tempo slow

$\bar X_{B}=21.17$ represent the mean for music tempo fast

$s_{A}=15.90$ represent the sample standard deviation for music tempo slow

$s_{B}=1.3$ represent the sample standard deviation for music tempo fast

$n_{A}=19$ sample size for the group 2

$n_{B}=14$ sample size for the group 2

$\alpha=0.10$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the true mean is higher when the music is slow, the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 0$

Alternative hypothesis: $\mu_{A} - \mu_{B} 0$

We don't have the population standard deviation's but we assume that the population deviation is equal for both populations, so we can apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{s_p\sqrt{\frac{1}{n_{A}}+\frac{1}{n_{B}}}}$ (1)

Where s_p represent the standard deviation pooled given by:

$s_p =\sqrt{\frac{(n_A -1)s^2_A +(n_B -1)s^2_B}{n_A +n_B -2}}$

$s_p =\sqrt{\frac{(19 -1)(15.9)^2 +(14-1)(9.1)^2}{19 +14 -2}}=13.473$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(30.44-21.17)-0}{13.473\sqrt{\frac{1}{19}+\frac{1}{14}}}}=1.9534$

P value

We need to find first the degrees of freedom given by:

$df=n_A +n_{B}-2=19+14-2=31$

Since is a one right tailed test the p value would be:

$p_v =P(t_{31}1.9534)=1-P(t_{31}$

Answer from: Quest

Where the answer choices?

Answer from: Quest

It is 2 out of 8 she will pick green and 3 out of 7 she will pick green but red and green would be 5 out of 5 so 50% chance

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