Help on numbers 4 and 6 please. Writing a function rule

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 436.0 gram setting. It

Vilka [71]

Answer:

We conclude that the machine is under filling the bags.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 436.0 gram

Sample mean, $\bar{x}$ = 429.0 grams

Sample size, n = 40

Alpha, α = 0.05

Population standard deviation, σ = 23.0 grams

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 436.0\text{ grams}\\H_A: \mu < 436.0\text{ grams}$

We use one-tailed(left) z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{429 - 436}{\frac{23}{\sqrt{40}} } = -1.92$

Now, $z_{critical} \text{ at 0.05 level of significance } = -1.64$

Since,

$z_{stat} < z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis. Thus, we conclude that the machine is under filling the bags.

8 0

3 years ago

A psychological experiment was conducted to investigate the length of time (time delay) between the administration of a stimulus

aleksklad [387]

Answer:

The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.

As there is significant evidence to reject $H_0: \mu= 1.6$ , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.

The significance level for this test is 0.05.

Step-by-step explanation:

In this case, we want to prove if there is significant evidence that the mean differs from 1.6 seconds. That is the same as having evidence to reject the the hypothesis $H_0:\mu= 1.6$ (the null hypothesis always have the equal sign).

We have to test the null hypothesis

$H_0:\mu= 1.6$

The significance level for this test is 0.05

Calculation of the t-statistic:

$t=\frac{M-\mu}{s/\sqrt{n}} =\frac{2.2-1.6}{0.57/\sqrt{36}} =\frac{0.6}{0.063}= 9.47$

If we look up in a t-table for t=9.47 and df=(36-1)=35, we get this value appears with a probability of zero. A large number like that is very unlikely to happen.

The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.

As there is significant evidence to reject $H_0: \mu= 1.6$ , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.

6 0

3 years ago

In a survey, 205 people indicated they prefer cats, 160 indicated they prefer dots, and 40 indicated they don’t enjoy either pet

maw [93]

Answer: probability = 0.506

Step-by-step explanation:

The data we have is:

Total people: 205 + 160 + 40 = 405

prefer cats: 205

prefer dogs: 160

neither: 40

The probability that a person chosen at random prefers cats is equal to the number of people that prefer cats divided the total number of people:

p = 205/405 = 0.506

in percent form, this is 50.6%

6 0

3 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

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 normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$