Detailed Answer

a) The probability is calculated using the normal cumulative distribution function:

\[ \text{normCdf}(250, 999, 248, 3) = 0.252 \]

b) The mean remains the same, regardless of the sample size. However, the standard deviation changes as follows:

\[ \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{10}} \]

Using this in our calculator:

\[ \text{normCdf}(0, 247, 250, \frac{3}{\sqrt{10}}) = 0.000783 \]

Detailed Answer

a) Here we simply need to enter the data into the z-interval section of our calculator to find, \[ 42.3 < \mu < 42.9\]

b) A mean of 40 does lie not within the 95% confidence interval, therefore we can reject the claim that it is 40.

Detailed Answer

a) \(s_{n-1} = \sqrt{\frac{n}{n-1}} \cdot s_n = \sqrt{\frac{30}{30-1}} \cdot 10 = 10.2\)

b) We need to use t values, as we don't know the population's variance.

\[ \text{Error} = t_{v,\frac{\alpha}{2}} \cdot \frac{s_{n-1}}{\sqrt{n}} \] \[ = t_{29,0.025} \cdot \frac{10.2}{\sqrt{30}} \] \[ = 2.06 \cdot \frac{10.2}{\sqrt{30}} \] \[ = 3.84 \]

Therefore, \[ 1532 - 3.84 < \mu < 1532.22 + 3.84 \] \[ 1528.16 < \mu < 1535.84 \]

c) Since the mass provided by the manufacturer is within the 95% confidence interval, we accept the manufacturer's claim.

Detailed Answer

a) We can input these values into our GDC and use One-variable statistics to find the values. \( \mu \) is straightforward to find as that is the average. We are also asked to estimate the variance, which we can do by squaring the standard deviation provided by the calculator: \( \sigma^2 = 2.13^2 = 4.52 \).

b) Here we need to use t-distribution, as the population variance is unknown. We also know \( \alpha = 0.05 \) and \( v = 9 \).

\[ \text{Error} = t_{v,\frac{\alpha}{2}} \cdot \frac{s}{\sqrt{n}} \] \[ = \text{invT}(0.975,9) \cdot \frac{\sqrt{4.52}}{\sqrt{10}} \] \[ = 1.52 \]

Thus, the required interval is: \[ 60.95 - 1.52 < \mu < 60.95 + 1.52 \] \[ 59.43 < \mu < 62.47 \]

c) (i) We can continue the same spreadsheet as what we started in the previous part and conduct a t-test for the data with our GDC. We find \( t = 1.41 \) and \( p\text{-value} = 0.19 \).

c) (ii) Since our p-value is larger than \( 0.05 \), we do not reject \( H_0 \), meaning that the initial assumption of 60 minutes was correct.

Detailed Answer

a) The mean is calculated as follows: \[ \text{Mean} = \frac{\sum x}{n} = \frac{1954}{20} = 97.7 \]

b) The sample standard deviation is calculated using the formula: \[ s_{n-1} = \sqrt{\frac{191100 - \frac{1954^2}{20}}{20-1}} = 3.2 \]

c) We can calculate the t-interval with our GDC, under the statistical tests subsection. Thus, we find: \[ 95.7 < \mu < 99.7 \]

d) Since the manufacturer claimed 100 chips per bag, which is not inside our 99% confidence interval, we can conclude that the manufacturer's claim is incorrect.

Detailed Answer

a) Here we can use Z-distribution as we know the population standard deviation. We know \(1 - \alpha = 0.8\) so \(\frac{\alpha}{2} = 0.1\). So the error, \[ \text{Error} = Z_{0.1} \cdot \frac{\sigma}{\sqrt{n}} = \text{normInv}(0.9) \cdot \frac{5}{\sqrt{50}} = 1.28 \cdot \frac{5}{\sqrt{50}} = 0.9 \]

Thus the required interval is, \[ 82 - 0.9 < \mu < 82 + 0.9 \] \[ 81.1 < \mu < 82.9 \]

b) \[ Z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{n}} = \frac{1}{2}\left( Z_{\frac{\alpha}{2}} \cdot \frac{\sigma}{\sqrt{50}} \right) \] \[ \sqrt{\frac{1}{n}} = \frac{1}{2\sqrt{50}} \] \[ n = 200 \]

Detailed Answer

a) \[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \overline{x})^2 = \frac{1}{19} \cdot 350 = 18.4 \]

b) (i) We know that the mean is at the mid-point of the provided interval.

\[ \overline{x} = \frac{322.87 + 331.45}{2} = 327.16 \]

Since we do not know \( \sigma \), we must use the t-distribution on our calculator. From which we can find that the confidence interval is:

\[ 325.151 < \mu < 329.169 \]

b) (ii) We can write the general form of the confidence interval as follows: \[ \overline{x} - t_{n-1, \frac{\alpha}{2}} \frac{s}{\sqrt{n}} \le \mu \le \overline{x} + t_{n-1, \frac{\alpha}{2}} \frac{s}{\sqrt{n}} \] \[ 327.16 - t_{19, \frac{\alpha}{2}} \frac{\sqrt{18.4}}{\sqrt{20}} \le \mu \le 327.16 + t_{19, \frac{\alpha}{2}} \frac{\sqrt{18.4}}{\sqrt{20}} \]

We know that the lower bound is 322.87 as we are told that in the question, so we can equate that with the expression we came up with: \[ 327.16 - t_{19, \frac{\alpha}{2}} \frac{\sqrt{18.4}}{\sqrt{20}} = 322.87 \] \[ -t_{19, \frac{\alpha}{2}} \frac{\sqrt{18.4}}{\sqrt{20}} = -4.29 \] \[ t_{19, \frac{\alpha}{2}} = 4.29 \frac{\sqrt{20}}{\sqrt{18.4}} = 4.47 \] \[ \frac{\alpha}{2} = P(T \ge t_{19, \frac{\alpha}{2}}) \] \[ \alpha = 2 \cdot 0.000131 = 0.000262 \] \[ a = 100 - 0.0262 = 99.97 \]

Detailed Answer

a) First, we need to calculate the changes in the distances jumped by the people.

We can then put this into our calculator and find that \[ \overline{x} = 3.8 \] \[ s_{n-1}^2 = 181 \]

b) (i) \( H_0: \mu_x = 0 \quad \text{and} \quad H_1: \mu_x > 0 \)

b) (ii) We can use our GDC again as we already have the data. We find \( t = 0.892. \)

b) (iii)

\[ t_{crit} = t_{v,\alpha} \] \[ = t_{9,0.05} = \text{invT}(0.95,9) = 1.83 \]

We can observe that \( t < t_{crit} \), meaning that there is not enough evidence suggesting that this 1 month period is enough to bring about a noticable increase of performance.

c) We know that the error is half the length of this interval: \[ t_{9,\frac{\alpha}{2}} \cdot \frac{s_{n-1}}{\sqrt{n}} = \frac{15.1 + 7.5}{2} = 11.3 \] \[ t_{9,\frac{\alpha}{2}} \cdot \frac{\sqrt{181}}{\sqrt{10}} = 11.3 \] \[ t_{9,\frac{\alpha}{2}} = 2.66 \] \[ 1 - \frac{\alpha}{2} = tCDF(-200, 2.66, 9) = 0.987 \] \[ 1 - \alpha = 0.974 \]

Thus, the confidence level is 97.4%.