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guajiro [1.7K]
2 years ago
13

An inspector inspects a shipment of medications to determine the efficacy in terms of the proportion p in the shipment that fail

ed to retain full potency after 60 days of production. Unless there is clear evidence that this proportion is less than 0.05, she will reject the shipment. To reach a decision, she will test the following hypotheses using the large-sample test for a population proportion:
H0 : p = 0.05, Ha : p < 0.05
To do so, she selects an SRS of 200 pills. Suppose that eight of the pills have failed to retain their full potency. The proportion 'p-hat' for the proportion of pills that have failed to retain their potency is:_____.
Mathematics
1 answer:
Dafna1 [17]2 years ago
4 0

Solution :

It is given that :

$H_0:p \geq 0.05$     (Null hypothesis, $H_0 : p = 0.05$ is also correct)

$H_a:p     (Alternate hypothesis, also called $H_1$)

This is a lower tailed test,

Therefore,

Sample proportion, $\hat p = \frac{x}{n}$

                                    $=\frac{8}{200}$

                                    = 0.04

And claimed proportion, P = 0.05

Significance level, $\alpha$ = 0.01

Therefore, calculating the statistics, we get

Standard deviation of $\hat p, \sigma_{\hat p}=\sqrt{\frac{P\times (1-P)}{n}}$

                                             $=\sqrt{\frac{0.05\times (1-0.05)}{200}}$

                                               $\approx 0.0154$

Test statistic,

$z_{observed}=\frac{\hat p -0.05}{\sigma_{\hat p}}$

              $=\frac{0.04 -0.05}{0.0154}$

               $\approx -0.65$

Now since this is a lower tailed test, p-value = $P(Z < z_{observed})=P(Z < -0.65)=0.2578$

Rejection Criteria : Reject $H_0$ if p-value < α .

Conclusion : Since the p-value $\geq \alpha$, we fail to reject the null hypothesis. There is insufficient evidence that p is less than 0.05

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The pulse rates of 174 randomly selected adult males vary from a low of 39 bpm to a high of 111 bpm. Find the minimum sample siz
Daniel [21]

Answer:

a. The sample size is approximately 97 adults

Step-by-step explanation:

The given parameters are;

The number of adults surveyed, n = 174

The lowest value of the pulse rate = 39 bpm

The highest value of the pulse rate = 111 bpm

The level of confidence used for determining the sample size = 90%

The given sample mean = 3 pm of the population mean

a. The range rule of thumb states that the standard deviation is approximately one quarter (1/4) of the range

The range = 111 bpm - 39 bpm = 72 bpm

Therefore, the standard deviation, σ = 72 bpm/4 = 18 bpm

The sample size, 'N', is given as follows;

N = \dfrac{z^2 \cdot p \cdot q}{e^2}

Where;

N = The sample size

z = The confidence level, 90% (z-score at 90% = 1.645)

p·q = σ² = 18² = 324

e² = 3² = 9

N = \dfrac{1.645^2  \times 324}{3^2} = 97.4169

Therefore the appropriate sample size, N ≈ 97 adults.

8 0
2 years ago
Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal prob
Novay_Z [31]

Answer:

a) By the Central Limit Theorem, it is approximately normal.

b) The standard error of the distribution of the sample mean is 1.8333.

c) 0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d) 0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours

e) 0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 36 hours and a standard deviation of 5.5 hours.

This means that \mu = 36, \sigma = 5.5

a. What can you say about the shape of the distribution of the sample mean?

By the Central Limit Theorem, it is approximately normal.

b. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)

Sample of 9 means that n = 9. So

s = \frac{\sigma}{\sqrt{n}} = \frac{5.5}{\sqrt{9}} = 1.8333

The standard error of the distribution of the sample mean is 1.8333.

c. What proportion of the samples will have a mean useful life of more than 38 hours?

This is 1 subtracted by the pvalue of Z when X = 38. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{38 - 36}{1.8333}

Z = 1.09

Z = 1.09 has a pvalue of 0.8621

1 - 0.8621 = 0.1379

0.1379 = 13.79% of the samples will have a mean useful life of more than 38 hours.

d. What proportion of the sample will have a mean useful life greater than 34.5 hours?

This is 1 subtracted by the pvalue of Z when X = 34.5. So

Z = \frac{X - \mu}{s}

Z = \frac{34.5 - 36}{1.8333}

Z = -0.82

Z = -0.82 has a pvalue of 0.2061.

1 - 0.2061 = 0.7939

0.7939 = 79.39% of the samples will have a mean useful life greater than 34.5 hours.

e. What proportion of the sample will have a mean useful life between 34.5 and 38 hours?

pvalue of Z when X = 38 subtracted by the pvalue of Z when X = 34.5. So

0.8621 - 0.2061 = 0.656

0.656 = 65.6% of the samples will have a mean useful life between 34.5 and 38 hours

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