Write a linear equation in standard form for the following scenario Matt is in charge of selling roses and chocolate hearts for the Valentine's Day dance he sell Beats Rose for $5 I need to Chocolate hard for $2.50 at the end of the dance he made a total of $250*

Answer:

Critical value:

The 90% confidence interval for the population mean bromide concentration is (0.376, 0.39).

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so . This value of z is the critical value

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the mean subtracted by M. So it is 0.383 - 0.0066 = 0.376 cc/cubic meter

The upper end of the interval is the mean added to M. So it is 0.383 + 0.0066 = 0.39 cc/cubic meter

The 90% confidence interval for the population mean bromide concentration is (0.376, 0.39).

Answer:

Mean = 5 feet 2 inches.

Step-by-step explanation:

1 feet = 12 inches

Height of Dad = 6 feet 2 inches  = {(6 × 12)+2} = 74 inches

Mom is 3 inches shorter than dad = 74 - 3 = 71 inches (5 ft 11 in)

Since mom is 2 inches taller than Ivan,

Height of Ivan = 71 - 2 = 69 inches  (5 ft 9 in)

Marica is 5 inches shorter than Ivan,

Height of Marica = 69 - 5 = 64 inches  (5 ft 4 in)

Marica is twice as tall as Sally-Jo.

Height of Sally-Jo = 64 ÷ 2 = 32 inches (2 ft 8 in)

Hence the mean height of the Schuller family

=

= 62 inches

converting 62 inches to feet = = 5 feet 2 inches

Mean height of the Schuller family is 5 feet 2 inches.

Answer:

a) 20

b) Equally likely, 0.05

Step-by-step explanation:

We are given the following in the question:

Population size, n = 6

Sample size, r = 3

a) Ways to select a sample of 3 from a population of 6

Thus, there are 20 ways in which a sample of 3 can be selected from a population of 6.

b) The chances that any particular sample will be the one selected are equally likely.

Probability of selecting one particular sample  =

Thus, 0.05 is the equally likely probability of selecting one sample.

Final answer:

A hypothesis test was conducted to evaluate the treatment's effect. For both variances, we failed to reject the null hypothesis, so we can't conclude that the treatment had a significant effect. The variability of scores plays a crucial role, as more variability makes it harder to identify a significant effect.

Explanation:

To determine if the treatment has a significant effect, we perform a hypothesis test using the sample mean (M), sample variance (s^2), and population mean (μ). The null hypothesis is that there's no effect from the treatment (μ=M), while the alternative hypothesis is that there is an effect (μ≠M).

a. For sample variance s^2=32, we can use the formula for the t score: t = (M - μ)/(s/√n) = (35 - 40)/(√32/√8) = -2.24. Based on a two-tailed t-distribution table, the critical t values for α=.05 and 7 degrees of freedom (n-1) are approximately -2.365 and 2.365. Our t value (-2.24) lies within this range, so we fail to reject the null hypothesis. We cannot conclude that the treatment has a significant effect.

b. Repeat the same process with sample variance s^2=72. The t value is now (35 - 40)/(√72/√8) = -1.48, again falling within the range of the critical t values. We can't conclude that the treatment has a significant effect.

c. As the variability (s^2) of the sample scores increases, it becomes more difficult to find a significant effect. Higher variability introduces more uncertainty, which can mask actual changes caused by the treatment.

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Final answer:

To evaluate the effect of a treatment using a two-tailed test with alpha = 0.05, we compare the calculated t-value to the critical t-value. The sample variance influences the outcome of the hypothesis test, with a larger variance leading to a wider critical region.

Explanation:

a. To test if the treatment has a significant effect, we will conduct a two-tailed hypothesis test using the t-distribution. The null hypothesis states that the treatment has no effect (μ = 40), while the alternative hypothesis states that the treatment has an effect (μ ≠ 40). With a sample size of 8, degrees of freedom (df) will be n-1 = 7. We will use the t-test formula to calculate the t-value, and compare it to the critical t-value from the t-table with α = 0.05/2 = 0.025. If the calculated t-value falls outside the critical region, we reject the null hypothesis and conclude that the treatment has a significant effect.

b. Similar to part a, we will conduct a two-tailed t-test using the same null and alternative hypotheses. With a sample size of 8, df = n-1 = 7. We will calculate the t-value using the sample mean, population mean, and sample variance. Comparing the calculated t-value to the critical t-value with α = 0.05/2 = 0.025, if the calculated t-value falls outside the critical region, we reject the null hypothesis and conclude that the treatment has a significant effect.

c. The variability of the scores in the sample, as indicated by the sample variance, influences the outcome of the hypothesis test. In both parts a and b, the sample variance is given. A larger sample variance (s^2 = 72 in part b) indicates more variability in the data, meaning the scores in the sample are more spread out. This leads to a larger t-value and a wider critical region. Therefore, it becomes easier to reject the null hypothesis and conclude that the treatment has a significant effect.

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