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koban [17]
1 year ago
11

(1 point) Since 1966, 45% of the No. 1 draftpicks in the NBA have been centers. identify the population A. all No.1 draftpick ce

nters in the NBA since 1966 B. all NBA players since 1966 C. all No.1 draftpicks in the NBA since 1966 D. None of the above identify the specified attribute A. being a No.1 draftpicks in the NBA B. being an NBA player C. being a center D. None of the above is the proportion 0.45 (45%) a population proportion or a sample proportion? A. population proportion B. sample proportion C. None of the above​
Mathematics
1 answer:
Firdavs [7]1 year ago
6 0

Answer:

1 B

2 A

3 B

Step-by-step explanation:

1

The population is the total sample from which part of the event is considered. It's like in a set, the population would encompass everything as a whole. From the statement given, we're expected to find the population. The population is B.

All NBA players since 1966. This is because it is from this pool that the draft picks were selected, and narrowed down to 45% which were centre backs.

2

The specified attribute is A.

Being a No. 1 draft pick, because that's the fulcrum of the statement in itself

3

B sample proportion. The population like I stated above, is all NBA players since 1966, thus, a population proportion would have had the statement like this.

45% of all NBA players since 1966...

So, therefore, it's a sample proportion since it is referring to the draft pick, and not the NBA players

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A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a st
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95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

                          P.Q. = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample credit hours per quarter = 15.2 credit hours

             s = sample standard deviation = 2.3 credit hours

             n = sample of students = 250

             \mu = population mean credit hours per quarter

<em>Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.</em>

So, 95% confidence interval for the population mean, \mu is ;

P(-1.96 < t_2_4_9 < 1.96) = 0.95  {As the critical value of t at 249 degree of

                                        freedom are -1.96 & 1.96 with P = 2.5%}  

P(-1.96 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-1.96 \times {\frac{s}{\sqrt{n} } } , \bar X+1.96 \times {\frac{s}{\sqrt{n} } } ]

                  = [ 15.2-1.96 \times {\frac{2.3}{\sqrt{250} } } , 15.2+1.96 \times {\frac{2.3}{\sqrt{250} } } ]

                  = [14.915 hours , 15.485 hours]

Therefore, 95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

<em>The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.</em>

8 0
1 year ago
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