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kow [346]
2 years ago
15

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks

that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.
Mathematics
1 answer:
kiruha [24]2 years ago
3 0

The standard normal distribution is a special type of normal distribution where the mean is 0, and the standard deviation is 1. The hypothesis can not be concluded.

<h3>What is Standard normal distribution?</h3>

The standard normal distribution is a special type of normal distribution where the mean is 0, and the standard deviation is 1.

For the given problems, the Hypothesis mean is 4.5. The sample mean was 4.75 hours with a normal standard deviation of 2.0.

Now, calculating the z-score,

z = \dfrac{\bar x - \mu_o}{\dfrac{\sigma}{\sqrt n}}\\\\\\z = \dfrac{4.75-4.5}{\dfrac{2}{\sqrt{15}}}\\\\\\z = 0.48

Now, as per the standard normal table, the p-value is 0.6844. Since the p-value is greater than the significance level, the hypothesis can not be concluded.

Hence, The hypothesis can not be concluded.

Learn more about Standard normal distribution:

brainly.com/question/25447725

#SPJ1

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Answer:

Step-by-step explanation:

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3 years ago
Please show your work!<br> 25% of 80
spin [16.1K]

Answer:

20

Step-by-step explanation:

25% change to .25

.25 x 80

Just solve 25 x 80 which equals 2000

then move decimal over twice giving you 20.00 or 20

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The service department of a luxury car dealership conducted research on the amount of time its service technicians spend on each
mart [117]

Answer:

Probability that the mean service time is between 1 and 2 hours is 0.96764.

Step-by-step explanation:

We are given that a systematic random sample of 100 service appointments has been collected.

The 100 appointments showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.

<u><em>Let </em></u>\bar X<u><em> = sample mean service time</em></u>

The z-score probability distribution for sample mean is given by;

                             Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = average preparation time = 90 minutes

           \sigma = standard deviation = 140 minutes

           n = sample of appointments = 100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, probability that the mean service time is between 60 and 120 minutes is given by = P(60 minutes < \bar X < 120 minutes)

P(60 minutes < \bar X < 120 minutes) = P(\bar X < 120 min) - P(\bar X \leq 60 min)  

  P(\bar X < 120 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{120-90}{\frac{140}{\sqrt{100} } } ) = P(Z < 2.14) = 0.98382

  P(\bar X \leq 60 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{60-90}{\frac{140}{\sqrt{100} } } ) = P(Z \leq -2.14) = 1 - P(Z < 2.14)

                                                        = 1 - 0.98382 = 0.01618

<em>The above probability is calculated by looking at the value of x = 2.14 in the z table which has an area of 0.98382.</em>

Therefore, P(60 min < \bar X < 120 min) = 0.98382 - 0.01618 = <u>0.96764</u>

7 0
3 years ago
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