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3 years ago

Which is bigger 0.305 or 0.350

Mathematics

2 answers:

kykrilka [37]3 years ago

6 0

Answer:

0.305

Step-by-step explanation:

When working with decimals, it would be the smaller number that is bigger. .305 is bigger because .305 is 305 points away from 0 and .350 is 350 points away from 0!!

Hope this helps!!!

-Unicorns110504

Katarina [22]3 years ago

3 0

Answer:

0.350 is bigger

Step-by-step explanation:

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A statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after

lana [24]

Answer:

95% confidence interval estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

(a) Lower Limit = 0.486

(b) Upper Limit = 0.624

Step-by-step explanation:

We are given that a statistician is testing the null hypothesis that exactly half of all engineers will still be in the profession 10 years after receiving their bachelor's.

She took a random sample of 200 graduates from the class of 1979 and determined their occupations in 1989. She found that 111 persons were still employed primarily as engineers.

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

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of persons who were still employed primarily as engineers = $\frac{111}{200}$ = 0.555

n = sample of graduates = 200

p = population proportion of engineers

Here for constructing 95% confidence interval we have used One-sample z proportion test statistics.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.555-1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ , $0.555+1.96 \times {\sqrt{\frac{0.555(1-0.555)}{200} } }$ ]

= [0.486 , 0.624]

Therefore, 95% confidence interval for the estimate for the proportion of engineers remaining in the profession is [0.486 , 0.624].

7 0

3 years ago

Do teachers find their work rewarding and satisfying? An article reports the results of a survey of 397 elementary school teache

Tju [1.3M]

Answer:

The 95% confidence interval would be given (0.0109;0.1651).

We are confident at 95% that the difference between the two proportions is between $0.0109 \leq p_{elementary} -p_{High school} \leq 0.1651$

Step-by-step explanation:

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".

$p_A$ represent the real population proportion for elementary school

$\hat p_A =\frac{226}{397}=0.569$ represent the estimated proportion for elementary school

$n_A=397$ is the sample size required for Brand A

$p_B$ represent the real population proportion for high school teachers

$\hat p_B =\frac{129}{268}=0.481$ represent the estimated proportion for high school teachers

$n_B=268$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.