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2 years ago

I’m having a bit issue with them can someone please help?

Mathematics

1 answer:

Snezhnost [94]2 years ago

5 0

You were on the right track. Here are two ways to go about it.

• By definition of conditional probability,

Pr[A | B] = Pr[A and B] / Pr[B]

Out of the total 100 participants in the survey, there are 18 people that both have a positive attitude and are over 35, so

Pr[positive and over 35] = 18/100

Out of the 100 pariticipants, 40 are over 35, so

Pr[over 35] = 40/100

Then the conditional probability you want is (18/100) / (40/100) = 18/40 = 9/20.

• There are 40 people over 35 in the survey. You want the probability that someone randomly chosen from this group has a positive attitude, of which there are 18. Hence the probability is 18/40 = 9/20.

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Answer:

$t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802$

The degrees of freedom are given by:

$df=n_{1}+n_{2}-2=4+4-2=6$

The p value for this case would be given by:

$p_v =2*P(t_{(6)}$

Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

Step-by-step explanation:

Information given

Battery 1 106 111 109 105

Battery 2 125 103 121 118

We can calculate the mean and the deviation with the following formulas"

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X_{1}=107.75$ represent the mean for the Battery 1

$\bar X_{2}=116.75$ represent the mean for the Bettery 2

$s_{1}=2.75$ represent the sample standard deviation for the Battery 1

$s_{2}=9.604$ represent the sample standard deviation for the battery 2

$n_{1}=4$ sample size selected for the Battery 1

$n_{2}=4$ sample size selected for the Battery 2

$\alpha=0.1$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to check if the difference in longevity between the two batteries, the system of hypothesis would be:

Null hypothesis: $\mu_{1} = \mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

The statistic is given by:

$t=\frac{\bar X_{s1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

The statistic is given by:

$t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802$

The degrees of freedom are given by:

$df=n_{1}+n_{2}-2=4+4-2=6$

The p value for this case would be given by:

$p_v =2*P(t_{(6)}$

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1,972
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