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hodyreva [135]
3 years ago
6

A researcher studying a memory enhancing technique had participants memorize nonsense words and definitions. Out of a list of 10

words, 57 participants were able to quickly learn 6.24 words on average with a standard deviation of 1.24. If it is known the mean number of words memorized in general is 5.65, is there sufficient evidence at that the memory enhancing technique helped participants learn more words?
a. What are the null and alternative hypotheses?
b. What is the value of the test statistic?
c. What is the critical value?
d. What is the decision?
e. What is the conclusion?
Mathematics
1 answer:
Setler [38]3 years ago
6 0

Answer:

A.

Null:

H0: u= 5.65

Alternative:

H1: u > 5.65

B.

The test statistic

Xbar = 6.24

u = 5.65

S = 1.24

N = 57

T = 6.24-5.65/(1.24/√57)

= 0.59/(1.24/7.5498)

= 0.59/0.1642

Test statistic = 3.593

C.

With alpha = 0.05

Df = n-1 = 57-1 = 56

T critical = 1.673

D.

If t test > 1.673 reject the null hypothesis

3.593>1.673, reject the null hypothesis

We conclude that there is enough evidence that the technique helped the participants to learn more.

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A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and
mr Goodwill [35]

Answer:

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

\mu = \frac{3X_{1} + 4X_{2}  }{8}

Now

Bias for the estimator = E(μ bar) - μ

                                    = E( \frac{3X_{1} + 4X_{2}  }{8}) - 4.5

                                    = \frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5

                                    = \frac{3(4.5) + 4(4.5)}{8} - 4.5

                                    = \frac{13.5 + 18}{8} - 4.5

                                    = \frac{31.5}{8} - 4.5

                                    = 3.9375 - 4.5

                                    = - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

                                                             = Var(μ bar) + [Bias(μ bar, μ)]²

                                                             = Var( \frac{3X_{1} + 4X_{2}  }{8}) + 0.3136

                                                             = \frac{1}{64} Var( {3X_{1} + 4X_{2}  }) + 0.3136

                                                             = \frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})]  }) + 0.3136

                                                             = \frac{1}{64} [{3(57.76) + 4(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [7(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [404.32]  } + 0.3136

                                                             = 6.3175 + 0.3136

                                                              = 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

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