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ira [324]
3 years ago
10

Use technology to help you test the claim about the population mean, mu, at the given level of significance, alpha, using the gi

ven sample statistics. Assume the population is normally distributed.
Claim: μ>1220;α=0.08; σ=211.67.
Sample statistics: x=1235.91,n=300
Identify the null and alternative hypotheses and calculate the standardized test statistic.
Mathematics
1 answer:
Sergio [31]3 years ago
3 0

Answer:

H0 : μ = 1220

H1 : μ > 1220

Test statistic = 1.30

Step-by-step explanation:

Sample mean, x = 1235.91

Standard deviation, σ = 211.67

Sample size, n = 300

The hypothesis :

Null ; H0 : μ = 1220

Alternative ; H1 : μ > 1220

Tbe test statistic :

(x - μ) ÷ (σ/√(n))

(1235.91 - 1220) ÷ (211.67/√(300))

15.91 / 12.220773

= 1.3018

= 1.30

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Answer:

Option B:

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Step-by-step explanation:

A function f(x) has a vertical asymptote if:

\lim_{x \to\\k^+}f(x) = \±\infty\\\\ \lim_{x \to\\k^-}f(x) = \±\infty

This means that if there is a value k for which f(x) has infinity or a -infinity then x = k is a vertical asymptote of f(x). Therefore, the closer x to k approaches, the closer the function becomes to infinity.

We can calculate the asymptote for function A.

\lim_{x \to \\1^+}(\frac{1}{x-1})\\\\ \lim_{x \to \\1^+}(\frac{1}{1^-1})\\\\ \lim_{x \to \\1^+}(\frac{1}{0}) = \infty\\\\and\\ \lim_{x \to \\1^-}(\frac{1}{x-1})\\\\\lim_{x \to \\1^-}(\frac{1}{0}) = -\infty

Then, function A has a vertical asymptote at x = 1.

The asymptote of function B can be easily observed in the graph. Note that the function b is not defined for x = -3 and when x is closest to -3, f(x) approaches infinity.

Therefore x = -3 is asintota of function B.

Therefore the correct answer is option B.

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