A game has 15 balls for each of the letters B, I, N, G, and O. The table shows the results of drawing balls 1,250 times.

1 answer:

4 0

Theoretically, each letter should have the same probability of occurring since there are 15 of each. There are 5 letters that can be drawn, so there is a total of 75 balls, and each letter has a probability $\dfrac{15}{75}=\dfrac15$ of being drawn.

This means one would expect a theoretical frequency of $\dfrac{1250}5=250$ , i.e. any given letter should get drawn 250 times, which means the answer is D.

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A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first sta

Akimi4 [234]

Answer:

95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

Step-by-step explanation:

We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).

In the first stages of a clinical trial, it was successful for 7 out of the 14 women.

Firstly, the pivotal quantity for 95% confidence interval for the true 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 women who find success with this new treatment = $\frac{7}{14}$ = 0.50

n = sample of women = 14

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

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

= [ $0.50-1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } }$ , $0.50+1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } }$ ]